Parabola

Given, equation of parabola

$ \begin{aligned} & y^2=16 x \\ & L L^{\prime}=4 a=16 \end{aligned} $

End points of latusrectum is $\left(a_1 \pm 2 a\right)$ Coordinates of $L$ is $(4,8)$ and $L^{\prime}$ is $(4,-8)$.

The end points of focal length is

$ \left(a t^2, 2 a t\right) \text { and }\left(\frac{a}{t^2},-\frac{2 a}{t}\right) $

Coordinate of $P$ is $(1,4)$

$ \begin{array}{ll} \Rightarrow & a t^2=1, \quad 2 a t=4 \\ \Rightarrow & 4 t^2=1, \quad 2 \times 4 \times t=4 \\ \Rightarrow & t^2=\frac{1}{4}, \quad t=\frac{1}{2} \\ \therefore & t=1 / 2 \end{array} $

∴ Coordinate of

$ Q\left(\frac{4}{\left(1 / 2^2\right.},-\frac{2 \times 4}{(1 / 2)}\right)=(16,-16) $

$ \begin{aligned} L Q & =\sqrt{(4-16)^2+(8+16)^2}=\sqrt{144+576} \\ & =\sqrt{720}=12 \sqrt{5} \end{aligned} $

Given, equation of parabola

$ y^2=32 x $

Here, $a=8$

The end point of focal chord is $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

$ \begin{array}{rlrl} \Rightarrow & a t^2 =\frac{1}{2} \text { and } 2 a t=4 \\ \Rightarrow & 8 \times t^2 =\frac{1}{2} \text { and } 2 \times 8 \times t=4 \\ \Rightarrow & t^2 =\frac{1}{16} \text { and } 16 t=4 \\ \Rightarrow & t= \pm \frac{1}{4} \text { and } t=\frac{1}{4} \\ & \therefore t =1 / 4 \end{array} $

Coordinate of $Q$ is $\left(\frac{8}{\left(\frac{1}{4}\right)^2}, \frac{-2 \times 8}{\frac{1}{4}}\right)$

$ =(128,-64) $

Coordinate of $S$ is $(8,0)$.

$ \begin{aligned} S Q & =\sqrt{(8-128)^2+(0+64)^2} \\ & =\sqrt{14400+4096}=\sqrt{18496}=136 \end{aligned} $

Let the point $P$ be $(h, k)$.

So, $\sqrt{(h-a)^2+(k-0)^2}=\frac{|h+k|}{\sqrt{1^2+1^2}}$

$ h^2+k^2+a^2-2 h a=\frac{h^2+k^2+2 h k}{2} $

$ 2 h^2+2 k^2+2 a^2-4 a h=h^2+k^2+2 h k $

∴ Locus of point $P$ is

$ x^2+y^2-2 x y-4 a x+2 a^2=0 $

Given, $y^2+4 y+8 x-2=0$

$ \begin{array}{r} y^2+4 y+4+8 x-6=0 \\ \left(y+2^2+2(4 x-3)=0\right. \\ y+2=0,4 x-3=0 \\ y=-2, x=3 / 4 \end{array} $

∴ Required point is $\left(\frac{3}{4},-2\right)$.

Statement I

Let $P(x, y)$ be any point on parabola. Distance from $P$ to focus = Perpendicular distance from $P$ to direction

$ \begin{aligned} & \sqrt{(x-2)^2+(y-3)^2}=\frac{|x+2 y+5|}{\sqrt{1^2+2^2}} \\ & 5\left(x^2+y^2-4 x-6 y+4+9\right)=x^2+4 y^2 +25+4 x y+20 y+10 x \\ & 4 x^2+y^2-4 x y-30 x-50 y+40=0 \end{aligned} $

∴ Statement I is true.

Statement II

$ \begin{aligned} x^2-4 x+16 y+52 & =0 \\ x^2-4 x+4 & =-16 y-48 \\ (x-4)^2 & =-16(y+3) \end{aligned} $

Comparison with $X^2=4 a Y$, we get vertex(2, - 3)

Equation of directrix is $Y=a$

$ y+3=4 \Rightarrow y=1 \text { or } y-1=0 $

∴ Statement II is wrong.

$ \begin{aligned} &\text { Given, } x=-2+2 t^2, y=2+4 t\\ &\begin{aligned} \Rightarrow & t=\frac{y-2}{4} \\ & x=-2+2\left(\frac{y-2}{4}\right)^2 \\ \Rightarrow & x=-2+\frac{y^2+4-4 y}{8} \\ & 8 x=-16+y^2+4-4 y \\ \Rightarrow & y^2-8 x-4 y-12=0 \end{aligned} \end{aligned} $

Given, equation is

$ \begin{array}{lr} & 2 x^2+5 y-6 x+1=0 \\ \Rightarrow & 2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0 \\ \Rightarrow & 2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right) \\ \Rightarrow & \left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right) \\ \Rightarrow & X^2=\frac{-5}{2} Y \end{array} $

where $X=x-\frac{3}{2}$ and $Y=y-\frac{7}{10}$

For vertex, on putting $X=0, Y=0$

$ \Rightarrow \quad x=3 / 2 \text { and } y=7 / 10 $

∴ Vertex $=(3 / 2,7 / 10)$

For focus, $X=0$ and $Y=-a$

$\Rightarrow x=\frac{3}{2}$ and $y-\frac{7}{10}=\frac{-5}{8}$

$\Rightarrow y=\frac{3}{40}$

∴ Focus $\equiv(3 / 2,3 / 40)$

As the axis of parabola is along the line $y=x$, the coordinates of focus and vertex are ( $h, h$ ) and ( $k, k$ ) respectively.

Distance of vertex from origin is $\sqrt{2}$.

$ \begin{array}{rlr} \Rightarrow & k^2+k^2 & =2 \\ \Rightarrow & k & =1 \end{array} $

Distance of focus from origin $=2 \sqrt{2}$

$ \begin{aligned} & & h^2+h^2 & =8 \Rightarrow h^2=4 \\ \Rightarrow & & h & =2 \end{aligned} $

Vertex $\equiv(1,1)$, focus $\equiv(2,2)$

Directrix passes through $(0,0)$ and perpendicular to $y=x$

∴ Directrix is $x+y=0$

∴ Parabola is

$ \begin{aligned} & (x-2)^2+(y-2)^2=\frac{(x+y)^2}{2} \\ & \Rightarrow \quad(x-y)^2+16=8(x+y) \end{aligned} $

which is satisfied by $x=(t+1)^2$,

$ y=(t-1)^2 $

Given, $y^2=16 x \Rightarrow a=4$

Let $B$ be $\left(x_1, 12\right)$ lies on parabola.

$ (12)^2=16 x_1 \Rightarrow x_1=\frac{12 \times 12}{16}=9 $

Point $B(9,12)$

Equation of focal chord passing through $(4,0)$ and ( 2,2 )

$ \begin{array}{cc} & y-0=\frac{2-0}{2-4}(x-4) \\ \Rightarrow & y=-x+4 \\ & x+y=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Equation of double ordinate :

$ x=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Point of intersection of Eqs. (i) and (ii), $(9,-5)$.

$ \begin{aligned}Given,\,\, & y^2=16 x \quad a=4 \\ & P(4,8) \equiv\left(a t_1^2, 2 a t_1\right) \\ & 2 a t_1=8 \\ & 8 t_1=8, \Rightarrow t_1=1 \end{aligned} $

If $P Q$ is a focal chord, then

$ \begin{aligned} t_1 t_2=-1 \Rightarrow+1 \times t_2=-1 & \\ t_2=-1 & \\ Q\left(a t_2^2, 2 a t_2\right) & \equiv Q(4,-8) \\ R(16,16) & \equiv\left(a t_3^2, 2 a t_3\right) \\ 2 a t_3 & =16 \\ 8 t_3 & =16 \Rightarrow t_3=2 \end{aligned} $

If $R T$ is a focal chord, then $t_3 t_4=-1$

$ \begin{aligned} & 2 \times t_4=-1 \Rightarrow t_4=\frac{-1}{2} \\ & T\left(a t_4^2, 2 a t_4\right) \equiv T(1,-4) \\ & Q T=\sqrt{3^2+4^2}=5 \end{aligned} $

From option (a).

$\begin{aligned} & x=4 \cos t, y=4 \sin t \\ & x^2=16 \cos ^2 t, y^2=16 \sin ^2 t \end{aligned}$

So, $x^2+y^2=16$, so this equation of a circle.

From option (b),

$x^2-2=-2 \cos t, y=\cos ^2\left(\frac{t}{2}\right)$

$ \begin{aligned} x^2 & =2-2 \cos t \quad \text{... (i)}\\ y & =\frac{1+\cos t}{2} \quad \text{... (ii)} \end{aligned}$

From Eq. (ii),

$\begin{aligned} \cos t= & 2 y-1 \\ \text { Now, } x^2 & =2-2 \cos t \\ x^2 & =2-2(2 y-1) \Rightarrow x^2=2-4 y+2 \\ x^2 & =4-4 y \end{aligned}$

So, this is represent the parabola.

Assume the vertex of the parabola is $(h, k)$ and the length of its latus rectum is $4a$. Since its axis is parallel to the $Y$-axis, the equation of the parabola can be written as:

$(x - h)^2 = 4a(y - k) \quad \text{.... (i)}$

The parabola passes through the points $(0, 4), (1, 9)$, and $(4, 5)$. Substituting these points into equation (i), we get:

$\begin{aligned} & \therefore (0 - h)^2 = 4a(4 - k) \\\\ & \Rightarrow h^2 = 4a(4 - k) \quad \text{... (ii)}\\\\ & (1 - h)^2 = 4a(9 - k) \\\\ & \Rightarrow 1 - 2h + h^2 = 4a(9 - k) \quad \text{... (iii)}\\\\ & (4 - h)^2 = 4a(5 - k) \\\\ & \Rightarrow 16 - 8h + h^2 = 4a(5 - k) \quad \text{... (iv)} \end{aligned}$

Next, we subtract Eqs. (ii) and (iii), and Eqs. (iii) and (iv), giving us:

$\begin{aligned} 1 - 2h & = 20a \quad \text{... (v)}\\\\ 15 - 6h & = -16a \quad \text{... (vi)} \end{aligned}$

By solving Eqs. (v) and (vi), we obtain $a = -\frac{3}{19}$ and $h = \frac{79}{38}$.

Substituting these values back, we find $k = \frac{9889}{912}$.

Therefore, the equation of the parabola becomes:

$\begin{aligned} & \left(x - \frac{79}{38}\right)^2 = \frac{-12}{19}(y - \frac{9889}{912}) \\\\ \Rightarrow & 19x^2 + 12y - 79x - 48 = 0 \end{aligned}$

Focus $\equiv(0,0)$

Tangent at vertex of parabola : $x-y+1=0$

The distance between the focus and tangent at the

$\text { vertex }=\left|\frac{0-0+1}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}}$

The directrix is the line parallel to the tangent at vertex and at a distance $2 \times \frac{1}{\sqrt{2}}$ from focus.

Let the equation of directrix be

$\begin{aligned} & x-y+\lambda=0 \\ & \text { Here, } \frac{\lambda}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}} \\ & \Rightarrow \quad \lambda=2 \end{aligned}$

$\therefore$ Equation of directrix is $x-y+2=0$

$\begin{aligned} & \text { Parabola : } y^2=4 p x \quad \text{... (i)}\\ & \text { Normal : } a x+b y=1 \quad \text{... (ii)}\\ & \Rightarrow \quad y=-\frac{a}{b} x+\frac{1}{b} \end{aligned}$

Slope, $m=-\frac{a}{b}$ and constant, $c=\frac{1}{b}$

Condition of normal says that

$\begin{aligned} & c=-2 m\left(\frac{\text { Coefficient of } x}{4}\right)-\left(\frac{\text { Coefficient of } x}{4}\right) m^3 \\ & \Rightarrow \frac{1}{b}=-2\left(\frac{-a}{b}\right)\left(\frac{4 p}{4}\right)-\left(\frac{4 p}{4}\right)\left(\frac{-a}{b}\right)^3 \\ & \Rightarrow \frac{1}{b}=\frac{2 p a}{b}+\frac{p a^3}{b^3} \\ & \Rightarrow b^2=2 p a b^2+p a^3 \\ & \Rightarrow p a^3=b^2-2 p a b^2 \end{aligned}$

Solving the given equations,

${(mx + c)^2} = 4ax - 4{a^3}$

$ \Rightarrow {m^2}{x^2} + 2mc\,.\,x + {c^2} = 4ax - 4{a^3}$

$ \Rightarrow {m^2}{x^2} + (2mc - 4a)x + {c^2} + 4{a^3} = 0$

Since the straight line touches the parabola at a point, so, the discriminant = 0

$ \Rightarrow {(2mc - 4a)^2} - 4{m^2}({c^2} + 4{a^3}) = 0$

$ \Rightarrow 4{m^2}{c^2} - 16amc + 16{a^2} - 4{m^2}{c^2} - 16{a^3}{m^2} = 0$

$ \Rightarrow - mc + a - {a^2}{m^2} = 0$

$ \Rightarrow mc = a - {a^2}{m^2} \Rightarrow c = {a \over m} - {a^2}m$

For the parabola ${y^2} = 4ax$, the equation of normal at $P(a{m^2}, - 2am)$ is $y = mx - 2am - a{m^3}$.

Here, $m = \tan 60^\circ = \sqrt 3 $

$\therefore$ $P \equiv (a{(\sqrt 3 )^2}, - 2a(\sqrt 3 )) \equiv (6, - 4\sqrt 3 )$ ($\because$ $a = 2$)

Thus, P satisfies $2x - y - 12 - 4\sqrt 3 = 0$

Suppose the parabola, y = x² + px + q cuts X-axis at A($\alpha$, 0) and B($\beta$, 0).

Then, $\alpha$, $\beta$ are roots of the equation x² + px + q = 0

$\therefore$ $\alpha$ + $\beta$ = $-$p and $\alpha \beta$ = q

The parabola y - x² + px + q cuts Y-axis at (0, q).

Let the equation of the circle passing through A, B and C be

x² + y² + 2gx + 2fy + c = 0 .... (i)

$\therefore$ $\alpha$² + 2g$\alpha$ + c = 0 ..... (ii)

$\beta$² + 2g$\beta$ + c = 0 ...... (iii)

and q² + 2fq + c = 0 ..... (iv)

Subtracting Eq. (iii) from Eq. (ii), we get

$\alpha + \beta + 2g = 0$

$\Rightarrow$ g = p/2

Adding Eqs. (ii) and (iii), we get

$\alpha^2+\beta^2+2g(\alpha+\beta)+2c=0$

$(\alpha+\beta)^2-2\alpha \beta+2g(\alpha+\beta)+2c=0$

p² $-$ 2q $-$ p² + 2c = 0 [$\because$ $\alpha+\beta=p$ and g = $\frac{p}{2}$]

Putting c = q in Eq. (iv), we get f = $-$$\left(\frac{q+1}{2}\right)$

Substituting the values of g, f and c in Eq. (i), we obtain the equation of family of circles passing through A, B and C as

x² + y² + px $-$ (q + 1) y + q = 0

Clearly, it passes through (0, 1).

Given, parabola

${y^2} = 8x$ ...... (i)

Equation of tangent at $P(2, - 4)$ is

$ - 4y = 4(x + 2)$

or, $x + y + 2 = 0$ ..... (ii)

and Equation of normal to the parabola is

$x - y + C = 0$

$\therefore$ Normal passes through $(2, - 4)$

$\therefore$ $C = - 6$

Normal : $x - y = 6$ ..... (iii)

Equation of directrix of parabola

$x = - 2$ ..... (iv)

Point of intersection of tangent and normal with directrix are $x = - 2$ at $A( - 2,0)$ and $B( - 2, - 8)$ respectively.

$Q(a,b)$ and $P(2, - 4)$ are given and AQBP is a square.

Mid-point of AB = Mid-point of PQ

$ \Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a = - 6,b = - 4$

$ \Rightarrow 2a + b = - 16$

Given parabola,

$y^2+4 x+2 y-8=0$ ..... (i)

Point of intersection of latusrectum and axis of parabola is foci.

Now, from Eq. (i), we get

$\begin{array}{ll} & (y+1)^2-1+4 x-8=0 \\ \Rightarrow \quad & (y+1)^2=-4 x+9 \\ \Rightarrow \quad & (y+1)^2=-4\left(x-\frac{9}{4}\right) \end{array}$

Compare with general standard form of parabola

where, foci $=(-a, 0)$

Here, $4 a=+4$

$\Rightarrow \quad a=+1$

Then, $\left(x-\frac{9}{4}\right)=-1 \Rightarrow x=-1+\frac{9}{4}=\frac{5}{4}$

and $(y+1)=0 \Rightarrow y=-1$

$\therefore \text { Focus }=\left(\frac{5}{4},-1\right)=\text { Point of intersection }$

$x=5t^2+2$ .... (i)

$y=10 t+4 \Rightarrow t=\frac{y-4}{10}$

From Eq. (i), $(x-2)=5\left(\frac{y-4}{10}\right)^2$

$\begin{array}{ll} \Rightarrow & (y-4)^2=20(x-2) \\ \Rightarrow & (7,4) \text { focus } \end{array} \quad\left\{\begin{array}{l} x-2=5 \\ y-4=0 \end{array}\right.$

Equation of parabola whose vertex is at origin and axis is $Y$-axis is

$x^2=4 a y$

This equation passes through $(6,-2)$

$\begin{aligned} & \therefore & 6^2 & =4 a(-2) \\ & \Rightarrow & \frac{-36}{8} & =a \Rightarrow a=\frac{-9}{2} \end{aligned}$

$\therefore$ Required parabola $x^2=-18 y$

Given parabola, $y^2=8 x$

Comparing with $y^2=4 a x$

We have, $a=2$

One end of focal chord $\left(\frac{1}{2}, 2\right)$

Parametric form of ends of focal chord $=\left(a t_1^2, 2 a t\right)$

$\begin{aligned} & \Rightarrow 2 a t=2 \\ & \therefore 2 \cdot 2 \cdot t=2 \Rightarrow t=\frac{1}{2} \end{aligned}$

$\begin{aligned} \text { Length of focal chord } & =a\left(t+\frac{1}{t}\right)^2 \\ & =2\left(\frac{1}{2}+2\right)^2=2 \cdot\left(\frac{5}{2}\right)^2=\frac{25}{2} \end{aligned}$

Let P(x, y) be the original position of the point w.r.t. the original axes. Let us move the origin at new position to (h, k).

Hence, the position of the same point P in the new system is

x' = x $-$ h $\Rightarrow$ y' = y $-$ k

Here, (h, k) = (1, 2)

$\therefore$ x' = (x $-$ 1), y' = (y $-$ 2)

According to the question,

y² $-$ 8x $-$ 4y + 12 = (y $-$ 2)² $-$ 4a(z $-$ 1)

$\Rightarrow$ y² $-$ 8x $-$ 4y + 12 = y² $-$ 4y + 4 $-$ 4ax + 4a

On comparing respective coefficients, we get

4a = 8 $\Rightarrow$ a = 2