Question

A focal chord of the parabola $y^2=16x$ touches the circle ${(x-6)}^2+y^2=2.$ Then the possible values of the slope of the chord are

• A. $\{-1,1\}$
• B. $\{-2,2\}$
• C. $\{-2,\frac{1}{2}\}$
• D. $\{2,-\frac{1}{2}\}$

Answer 1

The correct option is A $\{-1,1\}$

Focus of parabola $y^2=16x$ ...(1) is $(4,0)$
Thus equation of focal chord of (1) with slope say m is,
$(y-0)=m(x-4)\Rightarrow y=m(x-4)$ ...(2)
Now since line (2) touches circle $(x-6{)}^2+y^2=2$

So, its distance from centre of the circle is equal to the radius.
$\Rightarrow \left|\frac{m(6-4)}{\sqrt{1+m^2}}\right|=\sqrt{2}\to 2m^2=1+m^2\Rightarrow m=\pm 1$