Question

The slopes of the focal chords of the parabola $y^{2}=32 x$, which are tangents to the circle $x^{2}+y^{2}=4$, are

• A. $\frac{1}{2}, \frac{-1}{2}$
• B. $\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{15}}, \frac{-1}{\sqrt{15}}$
• D. $\frac{2}{\sqrt{5}}, \frac{-2}{\sqrt{5}}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{15}}, \frac{-1}{\sqrt{15}}$

Answer 2

Given equation of circle is $x^{2}+y^{2}=(2)^{2}$ $\therefore$ The equation of tangent to the circle is $y=m x \pm 2 \sqrt{1+m^{2}}\left(\because y=m x \pm r \sqrt{1+m^{2}}\right)$ Also, equation of parabola is $y^{2}=32 x$ $\therefore$ Focus of parabola is $(8,0)$ Since, the line passes through focus $(8,0)$. $\therefore 0=8 m \pm 2 \sqrt{1+m^{2}}$ $\Rightarrow -4 m=\pm \sqrt{1+m^{2}}$ $\Rightarrow 16 m^{2}=1+m^{2}$ $\Rightarrow 15 m^{2}=1$ $\Rightarrow m^{2}=\frac{1}{15}$ $\Rightarrow m=\pm \frac{1}{\sqrt{15}}$