Parabola

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

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

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 $=\left(a t_1^2, 2 a t\right)$

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

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

Given equation of parabola, $y^2=16 x$ and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola

Since, given parabola is symmetrical about $X$-axis

So, $\angle A O M=30^{\circ}$

In $\triangle A O M, \tan 30^{\circ}=\frac{A M}{O M}=\frac{8 t}{4 t^2}$

$ \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{2}{t} \Rightarrow t=2 \sqrt{3} $

∴ Coordinate of point $A$ lying on parabola is

$ \left(4(2 \sqrt{3})^2, 8(2 \sqrt{3})\right) \equiv(48,16 \sqrt{3}) $

∴ Distance

$ \begin{aligned} O A & =\sqrt{\left(x_A-x_0\right)^2+\left(y_A-y_0\right)^2} \\ & =\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2} \\ O A & =32 \sqrt{3} \end{aligned} $

∴ Length of the side of equilateral triangle $=32 \sqrt{3}$.

Given, equation of normal $m x-y+c=0$

Parabola $y^2=16 x$

Comparing with general equation of parabola i.e.,

$ y^2=4 a x \Rightarrow a=4 $

Let the coordinates of point $P$ on parabola be $\left(4 t^2, 8 t\right)$

∵ Slope of tangent at point $P=\frac{1}{t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, slope of tangent

$ \begin{equation*} =-\frac{1}{\text { Slope of normal }}=-\frac{1}{m} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

From Eqs. (i) and (ii), $\frac{1}{t}=-\frac{1}{m}$ or $t=-m$

or $t=-m$

∴ Coordinate of point is $P\left(4 m^2,-8 m\right)$.

Focus of parabola is $(a, 0)$ or $(4,0)$.

Given, focal distance $=40$

Using distance formula,

$ \begin{aligned} & \sqrt{\left(4 m^2-4\right)^2+(-8 m-0)^2}=40 \\ & \text { or } \sqrt{\left(m^2-1\right)^2+(-2 m)^2}=10 \\ & \text { or } \sqrt{\left(m^2-1\right)^2+4 m^2}=10 \end{aligned} $

$ \begin{aligned} & \text { or } \sqrt{\left(m^2+1\right)^2}=10 \\ & \text { or } m^2+1=10 \\ & \text { or } m^2=9 \\ & \text { or } m= \pm 3 \end{aligned} $

Therefore, coordinate of point $P$

$ \left(4 \times( \pm 3)^2,-8( \pm 3)\right) \equiv(36, \pm 24) $

∵ Point $P$ lies on line $m x-y+c=0$ then it must satisfy equation of line

$ \begin{aligned} & & ( \pm 3)(36)-( \pm 48)+c & =0 \\ & & \pm(3 \times 36+48)+c & =0 \\ \text { or } & & c & = \pm 132 \\ \text { or } & & |c| & =132 \end{aligned} $

$P Q$ is focal chord of parabola $y^2=4 x$ focus $S(1,0)$ and $P(4,4)$

We know that, Length of Semi-latus rectum is Harmonic mean of $P S$ and $S Q$

$ \begin{aligned} & \therefore \quad \frac{1}{P S}+\frac{1}{S Q}=1 \Rightarrow \frac{1}{S Q}=1-\frac{1}{P S} \\ & \frac{1}{S Q}=1-\frac{1}{\sqrt{9+16}}=1-\frac{1}{5}=\frac{4}{5} \Rightarrow S Q=\frac{5}{4} \end{aligned} $

$ \begin{gathered} x^2=4 a y \\ y=1+2 x \text { intersect the parabola } \\ P(0,1) P A=r_1, P B=-r_2 \\ \frac{x-0}{\frac{1}{\sqrt{5}}}=\frac{y-1}{\frac{2}{\sqrt{5}}}=r \end{gathered} $

$ \begin{aligned} \Rightarrow \quad x=\frac{r}{\sqrt{5}}, y & =\frac{2 r}{\sqrt{5}}+1 \\ x^2 & =4 a y \\ \Rightarrow \quad \frac{r^2}{5} & =4 a\left(\frac{2 r}{\sqrt{5}}+1\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad r^2-8 \sqrt{5} a r-20 a=0 \\ & \Rightarrow \quad-r_1 r_2=-20 a \end{aligned} $

$ \begin{aligned} & \Rightarrow & r_1 r_2=20 a \Rightarrow r_1-r_2 & =8 \sqrt{5} a \\ & & \left(r_1+r_2\right)^2-\left(r_1-r_2\right)^2 & =4 r_1 r_2 \\ & & 40-64 \times 5 a^2 & =80 a \\ & \Rightarrow & 8 a^2+2 a-1 & =0 \\ & & (4 a-1)(2 a+1) & =0 \\ & \Rightarrow & 4 a & =1 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } y=\frac{h^3}{3} x^2+\frac{h^2}{2} x-h+\frac{3}{4 h^3} \\ & 12 h^3 y=4 h^6 x^2+6 h^5 x-12 h^4+9 \\ & 4 h^6 x^2+6 h^5 x=12 h^3 y+12 h^4-9 \\ & \begin{aligned} \left(x^2+\frac{3}{2 h} x\right) & =\frac{12 h^4}{4 h^6}\left(y+\frac{12 h^4}{12 h^3}-\frac{3}{14 h^3}\right) \\ \left(x^2+\frac{3}{2 h^2}+\right. & \left.\frac{3}{16 h^2}\right) \\ & =\frac{3}{h^3}\left(y+h-\frac{3}{4 h^3}\right)+\frac{9}{16 h^2} \end{aligned} \end{aligned} \end{aligned} $

$ \left(x+\frac{3}{2 h}\right)^2=\frac{3}{h^3}\left(y+\frac{19 h}{16}-\frac{3}{4 h^3}\right) $

Here, $a=\frac{3}{4 h^3}$

∴ Equation of directrix is

$ \begin{aligned} & y+\frac{19 h}{16}-\frac{3}{4 h^3}=\frac{-3}{4 h^3} \Rightarrow y=-\frac{19 h}{16} \\ & \therefore \quad k=-\frac{19 h}{16} \Rightarrow \frac{k}{h}=-\frac{19}{16} \\ & \therefore \quad k: h=-19: 16 \end{aligned} $

Given equation of parabola

$ x^2=108 y(4 a=108 \Rightarrow a=27) $

Equation of tangent, $y=m x-27 m^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And equation of parabola

$ y^2=32 x(4 a=32 \Rightarrow a=8) $

Equation of tangent, $y=m x+\frac{8}{m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Since Eqs. (i) and (ii) are identical

$ \begin{aligned} & \therefore \quad m x-27 m^2=m x+\frac{8}{m} \\ & \Rightarrow \quad-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3} \end{aligned} $

∴ Required equation of tangent

$ \begin{array}{cc} & y=\left(\frac{-2}{3}\right) x-27\left(\frac{-2}{3}\right)^2 \\ \Rightarrow & y=\frac{-2 x}{3}-12 \\ \Rightarrow & 3 y=-2 x-36 \\ \Rightarrow & 2 x+3 y+36=0 \end{array} $

We have, equation of parabola is

$ \begin{array}{cr} & y^2+2 x+2 y-3=0 \\ \Rightarrow & (y+1)^2-1+2 x-3=0 \\ \Rightarrow & (y+1)^2=4-2 x \\ \Rightarrow & (y+1)^2=-2(x-2) \\ \therefore & \text { Vertex }=(2,-1) \end{array} $

Axis : $y+1=0$

Focus: $(-a+2,-1)=\left(-\frac{1}{2}+2,-1\right)$

                                         $ =\left(\frac{3}{2},-1\right) $

Directrix : $x-2=\frac{1}{2} \Rightarrow x=\frac{5}{2}$

$ \Rightarrow 2 x-5=0 $

We have, $y^2=4 x$

We know that, the centroid of the triangle formed by the conormal points on a parabola lies on its axis and its coordinates are $\left(\frac{2}{3}(h-2 a), 0\right)$, where $(h, k)$ is the point through which all normals passes. $\quad[\therefore a=1]$ ∴ Coordinate of centroid $=\left(\frac{2}{3}(5-2 \times 1), 0\right)=(2,0)$