Statement-1 : An equation of a common tangent to these curves is $y = x + \sqrt 5 $.
Statement-2 : If the line, $y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$ is their common tangent, then $m$ satiesfies ${m^4} - 3{m^2} + 2 = 0$.
$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$ is :
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)
$\begin{aligned} & \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\ & \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\ & \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\ & (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\ & \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\ & =192 \end{aligned}$
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
STATEMENT-2: A parabola is symmetric about its axis.
The ratio of the areas of the triangles $PQS$ and $PQR$ is
The radius of the circumcircle of the triangle $PRS$ is
The radius of the incircle of the triangle $PQR$ is
