Parabola

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 $=\left(\frac{2}{3}(5-2 \times 1), 0\right)=(2,0)$