Question

A circle of radius $4$ , drawn on a chord of the parabola $y^{2}=8 x$ as diameter, touches the axis of the parabola. Then, the slope of the chord is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. 1
• D. 2

Answer 1

Answer: (C)

Given, equation of parabola is
$y^{2}=8 x$ ...(i)
$\therefore a=2$
Let $(h, 4)$ be the coordinate of mid-point of chord.
Then, equation of chord is
$y-4=m(x-h)$ ...(ii)
If line (ii) passes through the point $P\left(2 t_{1}^{2}, 4 t_{1}\right)$ and $Q\left(2 t_{2}^{2}, 4 t_{2}\right)$ on parabola Eq. (i), then
$y\left(t_{1}+t_{2}\right)-2 x-4 t_{1} t_{2}=0$ ...(iii)
having slope
$m=\frac{2}{t_{1}+t_{2}}$ ...(iv)
Since, $(h, 4)$ is the mid-point of $P Q$. Therefore,
$2 \times 4=4\left(t_{1}+t_{2}\right)$
$\Rightarrow t_{1}+t_{2}=2$
Hence, slope of chord $P Q$ is
$m=\frac{2}{2}=1$ [using Eq. (iv)]