Question

The polar of any point on the circle $x^2+y^2-2ax-3a^2=0$ with respect to the circle $x^2+y^2+2ax-3a^2=0$ will touch the parabola:

• A. $x^2+4ay=0$
• B. $y^2+4ax=0$
• C. $y^2-4ax=0$
• D. None of the above

Answer 1

Answer: (B)

$y^2+4ax=0$

Let $(h,k)$ be any point on the circle $x^2+y^2-2ax-3a^2=0$

$\therefore h^2+k^2-2ah-3a^2=0$ ...[1]

Polar of $(h,k)$ w.r.t circle $x^2+y^2+2ax-3a^2=0$ is

$xh+ky+a(x+h)-3a^2=0$

$\therefore \left(\frac{-k}{h+a}\right)y=x-a\left(\frac{3a-h}{h+a}\right)$ ...[2]

Let $\frac{-k}{h+a}=t$ ...[3]

$\therefore t^2=\frac{k^2}{(h+a{)}^2}$

$=\frac{2ah+3a^2-h^2}{(h+a{)}^2}$ ...(from [1])

$=\frac{2ah+2a^2-h^2+a^2}{(h+a{)}^2}$

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

$=\frac{(3a-h)(h+a)}{(h+a{)}^2}$

$t^2=\frac{3a-h}{h+a}$ ....[4]

From [2], [3] and [4]

$ty=x-a{t}^2$

This line is the general equation of tangent to $y^2=-4ax$

So, the polar always touches the parabola $y^2+4ax=0$

So, the answer is option (B).