QJEEAdv15Maccb3: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Maccb3 Solution: If the normals of the parabola drawn at the end points of its…

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.