Parabola
Let $\mathrm{A}, \mathrm{B}$ and C be the vertices of a variable right angled triangle inscribed in the parabola $y^2=16 x$. Let the vertex $B$ containing the right angle be $(4,8)$ and the locus of the centroid of $\triangle A B C$ be a conic $C_0$. Then three times the length of latus rectum of $\mathrm{C}_0$ is $\_\_\_\_$
Explanation:
Since the right angle is at $B=(4,8)$, and $A,C$ lie on the parabola
$ y^2=16x, $
we first parametrize the parabola in the standard NCERT form.
For $y^2=4ax$, we have $4a=16 \Rightarrow a=4$.
So any point on the parabola can be written as
$ ( at^2,\, 2at )=(4t^2,\, 8t). $
Thus let
$ A=(4t_1^2,\, 8t_1), \qquad C=(4t_2^2,\, 8t_2). $
Also, the point $B=(4,8)$ lies on the parabola and corresponds to parameter $t=1$, since
$ (4\cdot 1^2,\, 8\cdot 1)=(4,8). $
So $B$ is the point with parameter $1$.
Now, for the triangle $ABC$ to be right-angled at $B$, the slopes of $BA$ and $BC$ must satisfy
$ m_{BA}\, m_{BC}=-1. $
For parabola $y^2=4ax$, the slope of chord joining parameter points $t_1$ and $t_2$ is
$ \frac{2a(t_2-t_1)}{a(t_2^2-t_1^2)}=\frac{2}{t_1+t_2}. $
So slope of chord joining parameter $1$ and $t_1$ is
$ m_{BA}=\frac{2}{1+t_1}, $
and slope of chord joining parameter $1$ and $t_2$ is
$ m_{BC}=\frac{2}{1+t_2}. $
Since these are perpendicular,
$ \frac{2}{1+t_1}\cdot \frac{2}{1+t_2}=-1. $
Hence
$ \frac{4}{(1+t_1)(1+t_2)}=-1 $
which gives
$ (1+t_1)(1+t_2)=-4. $
So,
$ t_1t_2+t_1+t_2+1=-4 $
or
$ t_1t_2+t_1+t_2=-5. $
Now let the centroid of $\triangle ABC$ be $G(h,k)$.
Using centroid formula,
$ h=\frac{4t_1^2+4+4t_2^2}{3} =\frac{4}{3}(t_1^2+t_2^2+1), $
$ k=\frac{8t_1+8+8t_2}{3} =\frac{8}{3}(t_1+t_2+1). $
Let
$ S=t_1+t_2, \qquad P=t_1t_2. $
Then from the perpendicularity condition,
$ P+S=-5. $
Also,
$ t_1^2+t_2^2=S^2-2P. $
So
$ h=\frac{4}{3}(S^2-2P+1). $
Since $P=-5-S$,
$ h=\frac{4}{3}\bigl(S^2-2(-5-S)+1\bigr) $
$ h=\frac{4}{3}(S^2+2S+11). $
Also,
$ k=\frac{8}{3}(S+1). $
From this,
$ S+1=\frac{3k}{8} \quad \Rightarrow \quad S=\frac{3k}{8}-1. $
Substitute into $h$:
$ h=\frac{4}{3}\left[\left(\frac{3k}{8}-1\right)^2+2\left(\frac{3k}{8}-1\right)+11\right]. $
Now simplify:
$ \left(\frac{3k}{8}-1\right)^2=\frac{9k^2}{64}-\frac{3k}{4}+1 $
and
$ 2\left(\frac{3k}{8}-1\right)=\frac{3k}{4}-2. $
So inside the bracket,
$ \frac{9k^2}{64}-\frac{3k}{4}+1+\frac{3k}{4}-2+11 = \frac{9k^2}{64}+10. $
Hence
$ h=\frac{4}{3}\left(\frac{9k^2}{64}+10\right) =\frac{3k^2}{16}+\frac{40}{3}. $
Therefore,
$ h-\frac{40}{3}=\frac{3k^2}{16}. $
Rearranging,
$ k^2=\frac{16}{3}\left(h-\frac{40}{3}\right). $
So the locus is a parabola:
$ k^2=4a\left(h-\frac{40}{3}\right) $
with
$ 4a=\frac{16}{3} \quad \Rightarrow \quad a=\frac{4}{3}. $
Length of latus rectum of this parabola is
$ 4a=\frac{16}{3}. $
Therefore, three times the length of latus rectum is
$ 3\times \frac{16}{3}=16. $
So the required answer is
$ \boxed{16}. $
Explanation:
$\begin{aligned} & (x-a)^2+(y-r)^2=r^2 \\ & (4-a)^2+(6-r)^2=r^2 \\ & 16+a^2-8 a+36+r^2-12 r=r^2 \\ & a^2-8 a-12 r+52=0 \end{aligned}$
Tangent to parabola at $(4,6)$ is
$6.4=9 .\left(\frac{x+4}{2}\right) \text { i.e. } 3 x-4 y+12=0$
This is also tangent to the circle
$\begin{aligned} & \therefore \quad C P=r \\ & \frac{3 a-4 r+12}{5}= \pm r \end{aligned}$
$3 \mathrm{a}+12=4 \mathrm{r} \pm 5 \mathrm{r}\left\{\begin{array}{l}\mathrm{ar} \\ -\mathrm{r}\end{array}\right.\quad\text{..... (1)}$
equation of circle is
$(x-a)^2+(y-r)^2=r^2$
satsty $P(4,6) \Rightarrow a^2-8 a-12 r+52=0\quad\text{..... (2)}$
From equation (1)
If $\mathrm{a}+4=3 \mathrm{r}$ then $\mathrm{a}=+6$ (rejected)
If $3 a+12=-r$ then $a=-14$ and $r=30$
Let $y^2=12 x$ be the parabola and $S$ be its focus. Let $P Q$ be a focal chord of the parabola such that $(S P)(S Q)=\frac{147}{4}$. Let $C$ be the circle described taking $P Q$ as a diameter. If the equation of a circle $C$ is $64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta$, then $\beta-\alpha$ is equal to $\qquad$ .
Explanation:
$\mathrm{y}^2=12 \mathrm{x} \quad \mathrm{a}=3 \quad \mathrm{SP} \times \mathrm{SQ}=\frac{147}{4}$
Let $\mathrm{P}\left(3 \mathrm{t}^2, 6 \mathrm{t}\right)$ and $\mathrm{t}_1 \mathrm{t}_2=-1$
(ends of focal chord)
So, $Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)$
$S(3,0)$
$\mathrm{SP} \times \mathrm{SQ}=\mathrm{PM}_1 \times \mathrm{QM}_2$
$\begin{aligned} & \text { (dist. from directrix) } \\ & =\left(3+3 \mathrm{t}^2\right)\left(3+\frac{3}{\mathrm{t}^2}\right)=\frac{147}{4} \\ & \Rightarrow \frac{\left(1+\mathrm{t}^2\right)^2}{\mathrm{t}^2}=\frac{49}{12} \\ & \mathrm{t}^2=\frac{3}{4}, \frac{4}{3} \\ & \mathrm{t}= \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}} \\ & \text { considering } \mathrm{t}=\frac{-\sqrt{3}}{2} \\ & \mathrm{P}\left(\frac{9}{4},-3 \sqrt{3}\right) \text { and } \mathrm{Q}(4,4 \sqrt{3}) \end{aligned}$
$\begin{aligned} &\text { Hence, diametric circle: }\\ &\begin{aligned} & (x-4)\left(x-\frac{9}{4}\right)+(y+3 \sqrt{3})(y-4 \sqrt{3})=0 \\ & \Rightarrow x^2+y^2-\frac{25}{4} x-\sqrt{3} y-27=0 \\ & \Rightarrow \alpha=400, \beta=1728 \\ & \beta-\alpha=1328 \end{aligned} \end{aligned}$
Let $A$ and $B$ be the two points of intersection of the line $y+5=0$ and the mirror image of the parabola $y^2=4 x$ with respect to the line $x+y+4=0$. If $d$ denotes the distance between $A$ and $B$, and a denotes the area of $\triangle S A B$, where $S$ is the focus of the parabola $y^2=4 x$, then the value of $(a+d)$ is __________.
Explanation:
$\begin{aligned} &\begin{aligned} \text { Area } & =\frac{1}{2} \times 4 \times 5=10=a \\ 6 & =4 \end{aligned}\\ &\text { So } \mathrm{a}+\mathrm{d}=14 \end{aligned}$
The focus of the parabola $y^2=4 x+16$ is the centre of the circle $C$ of radius 5 . If the values of $\lambda$, for which C passes through the point of intersection of the lines $3 x-y=0$ and $x+\lambda y=4$, are $\lambda_1$ and $\lambda_2, \lambda_1<\lambda_2$, then $12 \lambda_1+29 \lambda_2$ is equal to ________ .
Explanation:
$y^2=4(x+4)$
Equation of circle
$(x+3)^2+y^2=25$
Passes through the point of intersection of two lines $3 x-y=0$ and $x+\lambda y=4$ which is $\left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right)$, after solving with circle, we get
$\begin{aligned} & \lambda=-\frac{7}{6}, 1 \\ & 12 \lambda_1+29 \lambda_2 \\ & -14+29=15 \end{aligned}$
Let $A, B$ and $C$ be three points on the parabola $y^2=6 x$ and let the line segment $A B$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left(\frac{A M \cdot B N}{C D}\right)^2$ is equal to __________.
Explanation:
Equation of $A B$
$y\left(t_1+t_2\right)=2 x+2 a t_1 t_2$
$\begin{aligned} & \text { For } D, y=2 a t_3 \\ & \Rightarrow x=a\left(t_1 t_3+t_2 t_3-t_1 t_2\right) \\ & C D=\left|a\left(t_1 t_3+t_2 t_3-t_1 t_3\right)-a t_3^2\right| \\ & A M=\left|2 a t_1-2 a t_3\right| \\ & B N=\left|2 a t_3-2 a t_2\right| \\ & \left(\frac{A M \cdot B N}{C D}\right)^2=\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1 t_3+t_2 t_3-t_1 t_3-t_3^2\right)}\right)^2 \\ & =\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1-t_3\right)\left(t_2-t_3\right)}\right)^2 \\ & =16 a^2=16 \cdot\left(\frac{3}{2}\right)^2=36 \end{aligned}$
Consider the circle $C: x^2+y^2=4$ and the parabola $P: y^2=8 x$. If the set of all values of $\alpha$, for which three chords of the circle $C$ on three distinct lines passing through the point $(\alpha, 0)$ are bisected by the parabola $P$ is the interval $(p, q)$, then $(2 q-p)^2$ is equal to __________.
Explanation:
Chord with the middle point $(\alpha, 0)$
$\begin{aligned} & \Rightarrow T=S_1 \\ & \Rightarrow y y_1-4\left(x+x_1\right)=y_1^2-8 x_1 \\ & \Rightarrow-4(x+\alpha)=0-8 \alpha \\ & \Rightarrow x+\alpha=2 \alpha \Rightarrow x=\alpha \end{aligned}$
For circle chord with $(2 t^2, 4 t)$ as mid point
$\begin{aligned} & \Rightarrow \quad T=S_1 \\ & \Rightarrow \quad x x_1+y y_1-4=x_1^2+y_1^2-4 \\ & \Rightarrow \quad 2 t^2 x+4 t y=4 t^4+16 t^2 \end{aligned}$
Passes through $(\alpha, 0)$
$\begin{aligned} \Rightarrow & 2 t^2 \alpha=4 t^4+16 t^2 \\ \Rightarrow & 2 \alpha=4 t^2+16 \Rightarrow \alpha=2 t^2+8=x_0+8 \\ & x^2+y^2=4 \text { and } y^2=8 x \\ \Rightarrow & x^2+8 x-4=0 \Rightarrow x_0=\frac{-8+\sqrt{80}}{2} \\ \Rightarrow & p=8 \text { and } q=4+\frac{\sqrt{80}}{2} \Rightarrow(2 q-p)^2=80 \end{aligned}$
Let a conic $C$ pass through the point $(4,-2)$ and $P(x, y), x \geq 3$, be any point on $C$. Let the slope of the line touching the conic $C$ only at a single point $P$ be half the slope of the line joining the points $P$ and $(3,-5)$. If the focal distance of the point $(7,1)$ on $C$ is $d$, then $12 d$ equals ________.
Explanation:
As per given condition
$\begin{gathered} \frac{d y}{d x}=\frac{y+5}{2(x-3)} \\ \Rightarrow \ln (y+5)=\frac{1}{2} \ln (x-3)+c \\ \text { Passes through }(4,-2) \Rightarrow \ln 3=\frac{1}{2} \ln 1+c \\ \Rightarrow c=\ln 3 \end{gathered}$
$\Rightarrow$ Curve is $(y+5)^2=9(x-3)$
Focal distance of $(7,1)=\frac{9}{4}+4=\frac{25}{4}=d$
$12 d=75$
Let $L_1, L_2$ be the lines passing through the point $P(0,1)$ and touching the parabola $9 x^2+12 x+18 y-14=0$. Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\triangle P Q R$ is an isosceles triangle with base $Q R$. If the slopes of the lines $Q R$ are $m_1$ and $m_2$, then $16\left(m_1^2+m_2^2\right)$ is equal to __________.
Explanation:
$\begin{aligned} & 9 x^2+12 x+18 y-14=0 \\ & \left(x+\frac{2}{3}\right)^2=-2(y-1) \ldots(1) \end{aligned}$
Equation of tangent to (1)
$\begin{aligned} & t\left(x+\frac{2}{3}\right)=-(y-1)+\frac{1}{2} t^2 \text { passes through }(0,1) \\ & \Rightarrow \frac{2}{3} t=\frac{1}{2} t^2 \Rightarrow t=0, \frac{4}{3} \Rightarrow m=0, m=-\frac{4}{3} \end{aligned}$
$\begin{aligned} & \Rightarrow\left|\frac{m+\frac{4}{3}}{1-\frac{4}{3} m}\right|=|m| \Rightarrow\left|\frac{3 m+4}{3-4 m}\right|=|m| \\ & \Rightarrow 3 m+4=m(3-4 m) \text { or } 3 m+4=-m(3-4 m) \\ & 3 m+4=3 m-4 m^2 \text { or } 3 m+4=-3 m+4 m^2 \\ & 4 m^2+4=0 \text { (not possible) or } 4 m^2-6 m-4=0 \\ & m_1+m_2=\frac{3}{2}, m_1 m_2=-1 \\ & \Rightarrow m_1^2+m_2^2=\frac{17}{4} \\ & 16\left(m_1^2+m_2^2\right)=68 \end{aligned}$
Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point P. The distance of the point P from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is __________.
Explanation:
Line perpendicular to $2 x-y=10$ have slope $=\frac{-1}{2}$
$\Rightarrow$ Line tangent to parabola $y^2=4(x-9)$ with slope $m$ is
$\begin{aligned} & y=m(x-9)+\frac{1}{m}, m=\frac{-1}{2} \\ & \Rightarrow y=\frac{-(x-9)}{2}-2 \Rightarrow 2 y=-x+9-4 \\ & \Rightarrow 2 y+x=5 \end{aligned}$
Solving the tangent and parabola we get point $P$
$\begin{aligned} & \left(\frac{5-x}{2}\right)^2=4(x-9) \Rightarrow x^2-10 x+25=16 x-144 \\ & \Rightarrow x^2-26 x+169=0 \Rightarrow(x-13)^2=0 \\ & \Rightarrow P \equiv(13,-4) \end{aligned}$
Distance of $P$ from the centre of circle $(7,4)$ is $\sqrt{(13-7)^2+(-4-4)^2}=\sqrt{36+64}=10$ units.
Suppose $\mathrm{AB}$ is a focal chord of the parabola $y^2=12 x$ of length $l$ and slope $\mathrm{m}<\sqrt{3}$. If the distance of the chord $\mathrm{AB}$ from the origin is $\mathrm{d}$, then $l \mathrm{~d}^2$ is equal to _________.
Explanation:
Equation of focal chord
$y-0=\tan \theta .(x-3)$
Distance from origin
$\begin{aligned} & d=\left|\frac{-3 \tan \theta}{\sqrt{1+\tan ^2 \theta}}\right| \\ & I=4 \times 3 \operatorname{cosec}^2 \theta \\ & I. d^2=\frac{9 \tan ^2 \theta}{1+\tan ^2 \theta} \times 12 \operatorname{cosec}^2 \theta \\ & =\frac{108 \operatorname{cosec}^2 \theta}{1+\cot ^2 \theta}=108 \end{aligned}$
Let the length of the focal chord PQ of the parabola $y^2=12 x$ be 15 units. If the distance of $\mathrm{PQ}$ from the origin is $\mathrm{p}$, then $10 \mathrm{p}^2$ is equal to __________.
Explanation:
$\begin{aligned} & A B=15 \Rightarrow\left(3\left(t^2-\frac{1}{t^2}\right)\right)+\left(6\left(t+\frac{1}{t}\right)\right)^2=225 \\ & \Rightarrow 9\left(t^2-\frac{1}{t^2}\right)+36\left(t+\frac{1}{t}\right)^2=225 \end{aligned}$
$ \begin{aligned} \Rightarrow & \left.\left.\left(t+\frac{1}{t}\right)^2 \right[\,\left(t-\frac{1}{t}\right)+4\right]=25 \\ & \left(t+\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2=25 \Rightarrow\left(t+\frac{1}{t}\right)^4=25 \\ \Rightarrow & t+\frac{1}{t}= \pm \sqrt{5} \Rightarrow\left(t-\frac{1}{t}\right)= \pm 1 \end{aligned}$
Equation of $A B:(y-6 t)=\left(\frac{2 t}{t^2-1}\right)\left(x-3 t^2\right)$
$\Rightarrow$ Distance from $y-6 t=m x-3 m t^2$
$\Rightarrow p=\frac{\left|3 m t^2-6 t\right|}{\sqrt{1+m^2}}=\frac{\left|\left(\frac{6 t}{t^2-1}\right)\right|}{\sqrt{5}}=\frac{6}{\sqrt{5}}$
$\left[ m=\frac{2 t}{t^2-1}=\frac{}{t-\frac{1}{t}}= \pm 2 \Rightarrow m^2=4\right]$
$\Rightarrow \quad 10 p^2=\frac{10 \times 36}{5} \Rightarrow 72$
Explanation:
$x^2+y^2=3$ and $x^2=2 y$
$y^2+2 y-3=0 $$\Rightarrow(y+3)(y-1)=0$
$y=-3$ (Rejected) or $y=1$
For $\mathrm{y}=1, \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1)$
$p$ lies on the line
$ \begin{aligned} & \sqrt{2} x+y=\alpha \\\\ & \sqrt{2}(\sqrt{2})+1=\alpha \\\\ & \alpha=3 \end{aligned} $
For circle $\mathrm{C}_1$
$\mathrm{Q}_1$ lies on $\mathrm{y}$ axis
Let $\mathrm{Q}_1(0, \alpha)$ coordinates
$\mathrm{R}_1=2 \sqrt{3}$ (Given
Line $\mathrm{L}$ act as tangent
Apply P $=r$ (condition of tangency)
$\begin{aligned} & \Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3} \\\\ & \Rightarrow|\alpha-3|=6\end{aligned}$
$ \therefore $ $\alpha-3=6$
$\Rightarrow \alpha=9$
$\begin{gathered}\text { or } \alpha-3=-6 \\\\ \Rightarrow \alpha=-3\end{gathered}$
$\begin{aligned} & \triangle P Q_1 Q_2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1\end{array}\right| \\\\ & =\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2} \\\\ & \left(\triangle P Q_1 Q_2\right)^2=72\end{aligned}$
Let $P(\alpha, \beta)$ be a point on the parabola $y^2=4 x$. If $P$ also lies on the chord of the parabola $x^2=8 y$ whose mid point is $\left(1, \frac{5}{4}\right)$, then $(\alpha-28)(\beta-8)$ is equal to _________.
Explanation:
Parabola is $x^2=8 y$
Chord with mid point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{T}=\mathrm{S}_1$
$\begin{aligned} & \therefore \mathrm{xx}_1-4\left(\mathrm{y}+\mathrm{y}_1\right)=\mathrm{x}_1^2-8 \mathrm{y}_1 \\ & \therefore\left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(1, \frac{5}{4}\right) \\ & \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9 \end{aligned}$
$\therefore x-4 y+4=0$ ...... (i)
$(\alpha, \beta)$ lies on (i) & also on $y^2=4 x$
$\begin{aligned} & \therefore \alpha-4 \beta+4=0 \text{ .... (ii)} \\ & \& ~\beta^2=4 \alpha \text{ .... (iii)} \end{aligned}$
Solving (ii) & (iii)
$\begin{aligned} & \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\ & \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\ & \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\ & (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\ & \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\ & =192 \end{aligned}$
Explanation:
$\frac{d y}{d x}=\left.\frac{x}{-2 a} \Rightarrow \frac{d y}{d x}\right|_N=-t$
Slope of normal $=\frac{1}{t}=\frac{1}{\sqrt{6}} \Rightarrow t=\sqrt{6}$
Now, $\frac{-\mathrm{at}^2+\alpha}{2 \mathrm{at}}=\frac{1}{\mathrm{t}}$
$\Rightarrow-\mathrm{at}^2+\alpha=2 \mathrm{a}$
$\Rightarrow-6 \mathrm{a}+\alpha=2 \mathrm{a} \Rightarrow \alpha=8 \mathrm{a}$
For A and B
$\begin{aligned} & x^2=-4 a(-8 a) \\ & \Rightarrow x^2=32 a^2 \Rightarrow x= \pm 4 \sqrt{2} a \\ & \therefore A(-4 \sqrt{2} a,-8 a), B(4 \sqrt{2} a,-8 a) \\ & \therefore A B^2=(8 \sqrt{2} a)^2=128 a^2=s \\ & \therefore \text { Length of } L R=r=4 a \\ & \Rightarrow \frac{r}{s}=\frac{4 a}{128 a^2}=\frac{1}{16} \\ & \therefore 32 a=16 \Rightarrow a=\frac{1}{2} \\ & \therefore 24 a=12 \text { Ans. } \end{aligned}$
Let the tangent to the parabola $\mathrm{y}^{2}=12 \mathrm{x}$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$. Then the square of distance of the point $(6,-4)$ from the normal to the hyperbola $\alpha^{2} x^{2}-9 y^{2}=9 \alpha^{2}$ at its point $(\alpha-1, \alpha+2)$ is equal to _________.
Explanation:
$\Rightarrow \alpha= \pm 6$
$ \text { But, }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6 \text { reject }) $
Now, hyperbola $\frac{x^2}{9}-\frac{y^2}{36}=1$, normal at
$ \begin{aligned} & \mathrm{Q}(\alpha-1, \alpha+2) \text { is } \frac{9 \mathrm{x}}{5}+\frac{36 y}{8}=45 \\\\ & \Rightarrow 2 x+5 y-50=0 \end{aligned} $
Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to
$ \begin{aligned} & \left|\frac{2(6)-5(4)-50}{\sqrt{2^2+5^2}}\right|=\frac{58}{\sqrt{29}} \\\\ & \Rightarrow \text { Squareof distance }=116 \end{aligned} $
Let a common tangent to the curves ${y^2} = 4x$ and ${(x - 4)^2} + {y^2} = 16$ touch the curves at the points P and Q. Then ${(PQ)^2}$ is equal to __________.
Explanation:
$y^2=4 x$ is given by $y=m x+\frac{1}{m}$ and
Tangent of slope $m$ to the circle $(x-4)^2+y^2=16$ is given by
$ y=m(x-4) \pm 4 \sqrt{1+m^2} $
For common tangent
$ \begin{aligned} & \frac{1}{m}=-4 m \pm 4 \sqrt{1+m^2} \\\\ & \Rightarrow \left(\frac{1}{m}+4 m\right)^2=\left( \pm 4 \sqrt{1+m^2}\right)^2 \end{aligned} $
On squaring both sides, we get
$ \begin{aligned} & =\frac{1}{m^2}+16 m^2+8=16+16 m^2 \\\\ & \Rightarrow \frac{1}{m^2} =8 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}} \end{aligned} $
Then, the point of contact on parabola is $(8,4 \sqrt{2})$
Length of tangent $P Q$ from $(8,4 \sqrt{2})$ on the circle is
$ \begin{array}{ll} &\Rightarrow P Q=\sqrt{(8-4)^2+(4 \sqrt{2})^2-16} \\\\ &\Rightarrow P Q=\sqrt{16+32-16} \\\\ &\Rightarrow P Q=\sqrt{32} \Rightarrow(P Q)^2=32 \end{array} $
The ordinates of the points P and $\mathrm{Q}$ on the parabola with focus $(3,0)$ and directrix $x=-3$ are in the ratio $3: 1$. If $\mathrm{R}(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $\mathrm{P}$ and $\mathrm{Q}$, then $\frac{\beta^{2}}{\alpha}$ is equal to _______________.
Explanation:
Let the tangent to the curve $x^{2}+2 x-4 y+9=0$ at the point $\mathrm{P}(1,3)$ on it meet the $y$-axis at $\mathrm{A}$. Let the line passing through $\mathrm{P}$ and parallel to the line $x-3 y=6$ meet the parabola $y^{2}=4 x$ at $\mathrm{B}$. If $\mathrm{B}$ lies on the line $2 x-3 y=8$, then $(\mathrm{AB})^{2}$ is equal to ___________.
Explanation:
$x^2+2 x-4 y+9=0$ ..........(i)
Equation of tangent at $P(1,3)$ to the given curve (i)
$ \begin{array}{rlrl} & x(1)+2\left(\frac{x+1}{2}\right)-4\left(\frac{y+3}{2}\right)+9 =0 \\\\ & \Rightarrow 2 x+2 x+2-4 y-12+18 =0 \\\\ &\Rightarrow 4 x-4 y+8 =0 \\\\ &\Rightarrow x-y+2 =0 \end{array} $
which is meet the $Y$-axis at $A$
$\therefore A \equiv(0,2)$
Equation of line passing through $P$ and parallel to $x-3 y=6$ is $x-3 y+8=0$
Since, line (ii) meet the parabola $y^2=4 x$ at $B$
$ \begin{array}{lc} &\therefore y^2=4(3 y-8) \\\\ &\Rightarrow y^2-12 y+32=0 \\\\ &\Rightarrow (y-4)(y-8)=0 \end{array} $
$\therefore$ Possible co-ordinates of $B$ are $(4,4)$ and $(16,8)$.
Since, $(4,4)$ does not satisfies line $2 x-3 y=8$
Thus, $B$ is $(16,8)$
$ \therefore (A B)^2=(16-0)^2+(8-2)^2=256+36=292 $
If the $x$-intercept of a focal chord of the parabola $y^{2}=8x+4y+4$ is 3, then the length of this chord is equal to ____________.
Explanation:
Put $(3,0)$ in the above line $\mathrm{m}=-1$
Length of focal chord $=16$
Explanation:
As $\mathrm{P}(\mathrm{b}, \mathrm{c})$ lies on parabola so $\mathrm{c}^{2}=2 \mathrm{ab}---(1)$
Now equation of tangent to parabola $\mathrm{y}^{2}=2 \mathrm{ax}$ in point form is
$\mathrm{yy}_{1}=2 \mathrm{a} \frac{\left(\mathrm{x}+\mathrm{x}_{1}\right)}{2},$
Here, $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{b}, \mathrm{c})$
$\Rightarrow \mathrm{yc}=\mathrm{a}(\mathrm{x}+\mathrm{b})$
For point $\mathrm{B}$, put $\mathrm{y}=0$, now $\mathrm{x}=-\mathrm{b}$
So, area of $\triangle \mathrm{PBA}, \frac{1}{2} \times \mathrm{AB} \times \mathrm{AP}=16$
$ \begin{aligned} & \Rightarrow \frac{1}{2} \times 2 b \times c=16 \\\\ & \Rightarrow b c=16 \end{aligned} $
As $\mathrm{b}$ and $\mathrm{c}$ are natural number so possible values of $(b, c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$
Now from equation (1) $\mathrm{a}=\frac{\mathrm{c}^2}{2 \mathrm{~b}}$ and $\mathrm{a} \in \mathrm{N}$, so values of $(b, c)$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .
Hence sum of values of $a$ is 146 .
A triangle is formed by the tangents at the point (2, 2) on the curves $y^2=2x$ and $x^2+y^2=4x$, and the line $x+y+2=0$. If $r$ is the radius of its circumcircle, then $r^2$ is equal to ___________.
Explanation:
Tangent for ${y^2} = 2x$ at (2, 2) is
${L_1}:2y = x + 2$
Tangent for ${x^2} + {y^2} = 4x$ at (2, 2) is
${L_2}:y = 2$
${L_3}:x + y = 2 = 0$
Radius of circumcircle $ = {{abc} \over {4\Delta }}$
$ = {{(\sqrt {20} )(6)(\sqrt 8 )} \over {4 \times {1 \over 2} \times 6 \times 2}}$
$R = \sqrt {10} $
$R^2=10$
Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2,0)$ to the parabola $2 \mathrm{y}^{2}=-x$. If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x-5)^{2}+y^{2}=r$, then 17r is equal to ___________.
Explanation:
Given : ${y^2} = {{ - x} \over 2}$
$\eqalign{ & T \equiv y = mx - {1 \over {8m}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow (2,0) \cr} $
$ \Rightarrow {m^2} = {1 \over {16}} \Rightarrow m = \, \pm \,{1 \over 4}$
Tangents are $y = {1 \over 4}x - {1 \over 2},\,y = {{ - x} \over 4} + {1 \over 2}$
$4y = x - 2$ and $4y + x = 2$
If these are also tangent to circle then ${d_c} = r$
$ \Rightarrow \left| {{{5 - 2} \over {\sqrt {17} }}} \right| = \sqrt r \Rightarrow r = {\left( {{3 \over {\sqrt {17} }}} \right)^2}$
$ \Rightarrow 17r = 17\,.\,{9 \over {17}} = 9$
The sum of diameters of the circles that touch (i) the parabola $75 x^{2}=64(5 y-3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and (ii) the $y$-axis, is equal to ______________.
Explanation:
Equation of tangent to the parabola at $P\left(\frac{8}{5}, \frac{6}{5}\right)$
$ \begin{aligned} &75 x \cdot \frac{8}{5}=160\left(y+\frac{6}{5}\right)-192 \\\\ &\Rightarrow 120 x=160 y \\\\ &\Rightarrow 3 x=4 y \end{aligned} $
Equation of circle touching the given parabola at $\mathrm{P}$ can be taken as
$\left(x-\frac{8}{5}\right)^{2}+\left(y-\frac{6}{5}\right)^{2}+\lambda(3 x-4 y)=0$
If this circle touches $y$-axis then
$ \begin{aligned} &\frac{64}{25}+\left(y-\frac{6}{5}\right)^{2}+\lambda(-4 y)=0 \\\\ &\Rightarrow y^{2}-2 y\left(2 \lambda+\frac{6}{5}\right)+4=0 \\\\ &\Rightarrow D=0 \\\\ &\Rightarrow\left(2 \lambda+\frac{6}{6}\right)^{2}=4 \\\\ &\Rightarrow \lambda=\frac{2}{5} \text { or }-\frac{8}{5} \end{aligned} $
Radius $=1$ or 4
Sum of diameter $=10$
Let PQ be a focal chord of length 6.25 units of the parabola y2 = 4x. If O is the vertex of the parabola, then 10 times the area (in sq. units) of $\Delta$POQ is equal to ___________.
Explanation:
Given parabola ${y^2} = 4x$
$\therefore$ a = 1
Here, P, S, Q points are collinear.
$\therefore$ Slope of PS = Slope of QS
$ \Rightarrow {{2{t_1} - 0} \over {t_1^2 - 1}} = {{0 - 2{t_2}} \over {1 - t_2^2}}$
$ \Rightarrow {{2{t_1}} \over {t_1^2 - 1}} = {{2{t_2}} \over {t_2^2 - 1}}$
$ \Rightarrow {t_1}(t_2^2 - 1) = {t_2}(t_1^2 - 1)$
$ \Rightarrow t_2^2{t_1} - {t_1} = t_1^2{t_2} - {t_2}$
$ \Rightarrow t_2^2{t_1} - t_1^2{t_2} - {t_1} + {t_2} = 0$
$ \Rightarrow {t_1}{t_2}({t_2} - {t_1}) + ({t_2} - {t_1}) = 0$
$ \Rightarrow ({t_2} - {t_1})({t_1}{t_2} + 1) = 0$
As ${t_2} - {t_1} \ne 0$
$\therefore$ ${t_1}{t_2} + 1 = 0$
${t_1}{t_2} = - 1$
Now, lenght of PQ
$ = \sqrt {{{\left( {t_1^2 - t_2^2} \right)}^2} + {{\left( {2{t_1} - 2{t_2}} \right)}^2}} $
$ = \sqrt {{{\left( {{t_1} + {t_2}} \right)}^2}{{\left( {{t_1} - {t_2}} \right)}^2} + 4{{\left( {{t_1} - {t_2}} \right)}^2}} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2{t_1}{t_2} + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2( - 1) + 4} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2( - 1)} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2{t_1}{t_2}} $
$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} - {t_2}} \right)}^2}} $
$ = {\left( {{t_1} - {t_2}} \right)^2}$
Given, length of $PQ = {\left( {{t_1} - {t_2}} \right)^2} = 6.25$
$ \Rightarrow {t_1} - {t_2} = 2.5$
Now, Area of $\Delta OPQ$
$ = \left| {{1 \over 2}\left| {\matrix{ {t_1^2} & {2{t_1}} & 1 \cr {t_2^2} & {2{t_2}} & 1 \cr 0 & 0 & 1 \cr } } \right|} \right|$
$ = \left| {{1 \over 2}\left( {2{t_2}t_1^2 - 2{t_1}t_2^2} \right)} \right|$
$ = \left| {{1 \over 2} \times 2{t_1}{t_2}\left( {{t_1} - {t_2}} \right)} \right|$
$ = \left| {{t_1}{t_2} \times \left( {{t_1} - {t_2}} \right)} \right|$
$ = \left| { - 1 \times 2.5} \right|$
$ = 2.5$
$\therefore$ 10$\Delta$OPQ =
$ = 10 \times {{25} \over {10}} = 25$
A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola $y = {\left( {x - {1 \over 4}} \right)^2} + \alpha $, where $\alpha$ > 0. Then (4$\alpha$ $-$ 8)2 is equal to ______________.
Explanation:
Let the equation of circle be
$x\left( {x - {1 \over 2}} \right) + {y^2} + \lambda y = 0$
$ \Rightarrow {x^2} + {y^2} - {1 \over 2}x + \lambda y = 0$
Radius $ = \sqrt {{1 \over {16}} + {{{\lambda ^2}} \over 4}} = 2$
$ \Rightarrow {\lambda ^2} = {{63} \over 4}$
$ \Rightarrow {\left( {x - {1 \over 4}} \right)^2} + {\left( {y + {\lambda \over 2}} \right)^2} = 4$
$\because$ This circle and parabola $y - \alpha = {\left( {x - {1 \over 4}} \right)^2}$ touch each other, so
$\alpha = - {\lambda \over 2} + 2$
$ \Rightarrow \alpha - 2 = - {\lambda \over 2}$
$ \Rightarrow {(\alpha - 2)^2} = {{{\lambda ^2}} \over 4} = {{63} \over {16}}$
$ \Rightarrow {(4\alpha - 8)^2} = 63$
Let the common tangents to the curves $4({x^2} + {y^2}) = 9$ and ${y^2} = 4x$ intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then ${l \over {{e^2}}}$ is equal to ______________.
Explanation:
Let y = mx + c is the common tangent
So $c = {1 \over m} = \pm \,{3 \over 2}\sqrt {1 + {m^2}} \Rightarrow {m^2} = {1 \over 3}$
So equation of common tangents will be $y = \pm \,{1 \over {\sqrt 3 }}x \pm \,\sqrt 3 $, which intersects at Q($-$3, 0)
Major axis and minor axis of ellipse are 12 and 6.
So eccentricity
${e^2} = 1 - {1 \over 4} = {3 \over 4}$ and length of latus rectum $ = {{2{b^2}} \over a} = 3$
Hence, ${l \over {{e^2}}} = {3 \over {3/4}} = 4$
Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = ____________.
Explanation:
Focus = (4, 4) and vertex = (3, 2)
$\therefore$ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line x + 2y = 6 is B(x2, y2)
$\therefore$ ${{{x_2} - 2} \over 1} = {{{y_2} - 0} \over 2} = {{ - 2(2 + 0 - 6)} \over 5}$
$\therefore$ $B({x_2},\,{y_2}) = \left( {{{18} \over 5},{{16} \over 5}} \right)$.
Point B is point of intersection of direction with axes of parabola P2.
$\therefore$ $x + 2y = \lambda $ must have point $\left( {{{18} \over 5},{{16} \over 5}} \right)$
$\therefore$ $x + 2y = 10$
Explanation:
P(2, $-$4) $\Rightarrow$ $-$4 = 2m + ${2 \over m}$
$\Rightarrow$ m + ${1 \over m}$ = $-$2 $\Rightarrow$ m = $-$1
$\therefore$ tangent is y = $-$x $-$2
$\Rightarrow$ x + y + 2 = 0 ...... (1)
(1) is also tangent to x2 + y2 = a
So, ${2 \over {\sqrt 2 }} = \sqrt a \Rightarrow \sqrt a = \sqrt 2 $
$\Rightarrow$ a = 2
Explanation:
Normal at point P
$tx + y = 3t + {3 \over 2}{t^3}$
Passes through $\left( {3,{3 \over 2}} \right)$
$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$
$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$
$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$
$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$
Explanation:
focus : ($-$16, 0)
y = mx + c is focal chord
$\Rightarrow$ c = 16 m ...........(1)
y = mx + c is tangent to (x + 10)2 + y2 = 4
$\Rightarrow$ y = m(x + 10) $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ c = 10m $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ 16m = 10m $\pm$ 2$\sqrt {1 + {m^2}} $
$\Rightarrow$ 6m = 2$\sqrt {1 + {m^2}} $ (m > 0)
$\Rightarrow$ 9m2 = 1 + m2
$\Rightarrow$ m = ${1 \over {2\sqrt 2 }}$ & c = ${8 \over {\sqrt 2 }}$
$4\sqrt 2 (m + c) = 4\sqrt 2 \left( {{{17} \over {2\sqrt 2 }}} \right)$ = 34
Explanation:
Parabola : y2 = 4x
Let tangent y = mx + ${a \over m}$
y = mx + ${1 \over m}$
m2x $-$ my + 1 = 0
the above line is also tangent to circle
(x $-$ 3)2 + y2 = 9
$\therefore$ $ \bot $ from (3, 0) = 3
$\left| {{{3{m^2} - 0 + 1} \over {\sqrt {{m^2} + {m^4}} }}} \right| = 3$
(3m2 + 1)2 = 9(m2 + m4)
$6{m^2} + 1 + 9{m^4} = 9{m^2} + 9{m^4}$
$3{m^2} = 1$
$m = \pm {1 \over {\sqrt 3 }}$
$ \therefore $ tangent is
$y = {1 \over {\sqrt 3 }}x + \sqrt 3 $
(it will be used)
or
$y = - {1 \over {\sqrt 3 }}x - \sqrt 3 $
(rejected)
$m = {1 \over {\sqrt 3 }}$
For parabola
$\left( {{a \over {{m^2}}},{{2a} \over m}} \right) \equiv (3,2\sqrt 3 )$ = (c, d)
for circle $y = {1 \over {\sqrt 3 }}x + \sqrt 3 $
&
${(x - 3)^2} + {y^2} = 9$
Solving,
${(x - 3)^2} + {\left( {{1 \over {\sqrt 3 }}x + \sqrt 3 } \right)^2} = 9$
${x^2} + 9 - 6x + {1 \over 3}{x^2} + 3 + 2x = 9$
${4 \over 3}{x^2} - 4x + 3 = 0$
$4{x^2} - 12x + 9 = 0$
$4{x^2} - 6x - 6x + 9 = 0$
$2x(2x - 3) - 3(2x - 3) = 0$
$(2x - 3)(2x - 3) = 0$
$x = {3 \over 2}$
$ \therefore $ $y = {1 \over {\sqrt 3 }}\left( {{3 \over 2}} \right) + \sqrt 3 $
$y = {{\sqrt 3 } \over 2} + \sqrt 3 $
$y = {{3\sqrt 3 } \over 2}$
$(a,b) \equiv \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right)$
$2(a + c) = 2\left( {{3 \over 2} + 3} \right)$
$ = 2\left( {{3 \over 2} + {6 \over 2}} \right) = 9$
Explanation:
${{dy} \over {dx}} = {e^x}$
${\left. {{{dy} \over {dx}}} \right|_{x = c}} = {e^c}$
Tangent is $y - {e^c} = {e^c}(x - c)$
Put y = 0, x = c$ - $1.........(i)
For y2 = 4x
$2y{{dy} \over {dx}} = 4 \Rightarrow {\left. {{{ - dx} \over {dy}}} \right|_{y = 2}} = - 1$
Normal is $y - 2 = - 1(x - 1)$
Put y = 0, x = 3 ...........(ii)
From (i) and (ii); $c - 1$ = 3
$ \Rightarrow $ c = 4
Explanation:
let P(t2 , t)
Tangent at P(t2 , t)
ty = ${{x + {t^2}} \over 2}$
$ \Rightarrow $2ty = x + t2
$ \therefore $ Q = (-t2, 0)
Given ($\Delta $OPQ) = 4
${1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {{t^2}} & t & 1 \cr { - {t^2}} & 0 & 1 \cr } } \right|$ = 4
$ \Rightarrow $ $\left| {{t^3}} \right|$ = 8
$ \Rightarrow $ t = 2
and P = (4, 2)
As y = mx
$ \Rightarrow $ 2 = 4m
$ \Rightarrow $ m = 0.5
Explanation:
${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {5 \over 9} = {4 \over 9}$
The foci are ($\pm$ ae, 0) i.e. (2, 0) and ($-$2, 0).
The parabola P1 is ${y^2} = 8x$ and P2 is ${y^2} = - 16x$
As tangent with slope m1 to P1 passes through ($-$4, 0), we have
$y = {m_1}x + {2 \over {{m_1}}}$ giving $0 = - 4{m_1} + {2 \over {{m_1}}}$
i.e. $4m_1^2 = 2 \Rightarrow m_1^2 = {1 \over 2}$
Again for tangent with slope m2 to P2 passing through (2, 0), we have
$y = {m_2}x - {4 \over {{m_2}}} \Rightarrow 0 = 2{m_2} - {4 \over {{m_2}}}$
$ \Rightarrow 2m_2^2 = 4$ $\therefore$ $m_2^2 = 2$
Thus, ${1 \over {m_1^2}} + m_2^2 = 2 + 2 = 4$
Explanation:
Let, P(t2, 2t) be any point on the parabola y2 = 4x. C be the mirror image of the parabola y2 = 4x with respect to the line UV : x + y + 4 = 0.
The curve C cuts the line KL : y = $-$5 at A and B.
Let, B($\alpha$, $\beta$) be the image of the point P(t2, 2t).
Clearly, PB $\bot$ UV and PQ = QB.
$\therefore$ ${{\alpha - {t^2}} \over {\beta - 2t}} \times ( - 1) = - 1$
or, $\alpha - {t^2} = \beta - 2t$ ...... (1)
The point of intersection of the lines UV and KL is R.
Let us join P and R.
From $\Delta$PQR and $\Delta$BQR,
(i) BQ = PQ [$\because$ B is the image of P]
(ii) $\angle$PQR = $\angle$RQB = 90$^\circ$ [$\because$ PB $\bot$ UV]
(iii) QR common
$\therefore$ $\Delta$PQR $ \cong $ $\Delta$BQR [by SAS congruence criterion]
$\therefore$ $\angle$QRP = $\angle$BRQ [CPCT]
$\because$ slope of x + y + 4 = 0 is $-$1,
$\therefore$ $\angle$UTO = 135$^\circ$
$\therefore$ $\angle$OTR = 45$^\circ$
Again, X'X || KL and UV transversal.
$\therefore$ $\angle$OTR = $\angle$TRB = 45$^\circ$ $\therefore$ $\angle$BRQ = $\angle$QRP = 45$^\circ$
$\therefore$ $\angle$PRB = 90$^\circ$ $\therefore$ PR $\bot$ KL
$\therefore$ coordinates of R are (t2, $\beta$).
$\because$ the point R lies on KL,
$\therefore$ $\beta$ = $-$5
Again, the point R lies on the straight line x + y + 4 = 0.
$\therefore$ t2 + $\beta$ + 4 = 0
or, t2 $-$ 5 + 4 = 0 [$\because$ $\beta$ = $-$5]
or, t2 = 1 or, t = $\pm$ 1
when t = 1, $\beta$ = $-$5, then (1) $\Rightarrow$ $\alpha$ $-$ 1 = $-$ 5 $-$ 2 or $\alpha$ = $-$6
when, t = $-$1, $\beta$ = $-$5, then (1) $\Rightarrow$ $\alpha$ $-$ 1 = $-$ 5 + 2 or, $\alpha$ = $-$2
So, the coordinates of A and B are ($-$6, $-$5) and ($-$2, $-$5) respectively.
$\therefore$ AB = 4 units
So, the distance between A and B is 4 units.
Explanation:
Given: A parabola $y^2=4 x$
Comparing the given equation of parabola with the standard equation of parabola $y^2=4 a x$, we get $a=1$
Also, the end points of latus Rectum are $(a, \pm 2 a)$
$\Rightarrow$ The end points of latus rectum are $(1,2)$ and $(1,-2)$
Also we know that the equation of normal to the parabola at point
$\begin{aligned} & & \left(a m^2,-2 a m\right) \text { is } y & =m x-2 a m-a m^3 \\ \Rightarrow & & \left(a m^2,-2 a m\right) & =(1,2) \\ \Rightarrow & & \left(m^2,-2 m\right) & =(1,2) \\ \Rightarrow & & m^2 & =1 \text { and } m=-1 \\ \Rightarrow & & m & =-1 \end{aligned}$
So, the equation of the normal at $(1,2)$ is,
$\begin{aligned} & & y & =(-1) x-2(1)(-1)-(1)(-1)^3 \\ \Rightarrow & & y & =-x+3 \\ \Rightarrow & & x+y-3 & =0 \end{aligned}$
As the normal is tangent to the circle $(x-3)^2+ (y+2)^2=r^2$
$\Rightarrow$ The perpendicular distance of the tangent from the centre of the circle is equal to the radius of the circle.
Now, comparing the equation of the circle with the general form of the circle we get Coordinates of centre $\equiv(3,-2)$
$\Rightarrow$ Perpendicular distance from $(3,-2)$ to $x+y-3=r$
$\begin{array}{rrr} \Rightarrow & \left|\frac{3+(-2)-3}{\sqrt{1^2+1^2}}\right| & =r \\ \Rightarrow & \frac{2}{\sqrt{2}}=r \\ \Rightarrow & r^2=2 \end{array}$
Hint :
(i) The equation of the normal to the parabola at point $\left(a m^2,-2 a m\right)$ is $y=m x-2 a m-a m^3$.
(ii) The perpendicular distance of a point $(h, k)$ from the line $a x+b y+c=o$ is $\left|\frac{a h+b k+c}{\sqrt{a^2+b^2}}\right|$ units.
Explanation:
Given, circle is $x^2+y^2-2 x-4 y=0$ and parabola $y^2=8 x$.
$\because$ Both the curves intersect each other at P.
$\because \quad x^2+8 x-2 x-4 \cdot 2 \sqrt{2 x}=0$
$\begin{array}{lr} \Rightarrow & x^2+6 x-8 \sqrt{2 x}=0 \\ \Rightarrow & \sqrt{x}\left[x^{\frac{3}{2}}+6 x^{\frac{1}{2}}-8 \sqrt{2}\right]=0 \\ \Rightarrow & \text { Let } \sqrt{x}=t \\ \therefore & t\left[t^3+6 t-8 \sqrt{2}\right]=0 \\ \Rightarrow & t(t-\sqrt{2})\left(t^2-\sqrt{2} t+4\right)=0 \\ \Rightarrow & t=0 \text { or } t=\sqrt{2} \text { or } t=\frac{\sqrt{2} \pm \sqrt{2-4(4)}}{2} \end{array}$
(rejected because it is imaginary)
$\begin{array}{ll} \Rightarrow t=0 & \text { or } t=\sqrt{2} \\ \Rightarrow x=0 & \text { or } x=2 \\ \Rightarrow y=0 & \text { or } y=4 \end{array}$
Hence, the required coordinates are $\mathrm{P}(2,4), Q(0,0)$ and $S(2,0)$.
$\therefore \quad$ Area of $\triangle \mathrm{PQS}=\frac{1}{2} \times 2 \times 4=4$
Explanation:
The area of triangle formed by the three points on the parabola is twice the area of the triangle formed by the respective tangents. That is,
$\Delta LPM = 2 \times $ (Area of $\Delta ABC$)
${y^2} = 8x = 4 \times 2 \times x$
${{\Delta LPM} \over {\Delta ABC}} = 2$
${{{\Delta _1}} \over {{\Delta _2}}} = 2$
