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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
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{15}}, \frac{-1}{\sqrt{15}}$
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 $(8,0)$.
$$
\begin{aligned}
& \therefore \quad 0=8 m \pm 2 \sqrt{1+m^2} \\
& \Rightarrow \quad-4 m= \pm \sqrt{1+m^2} \\
& \Rightarrow \quad 16 m^2=1+m^2 \\
& \Rightarrow \quad 15 m^2=1 \\
& \Rightarrow \quad m^2=\frac{1}{15} \\
& \Rightarrow \quad m= \pm \frac{1}{\sqrt{15}} \\
&
\end{aligned}
$$
$$
x^2+y^2=(2)^2
$$
$\therefore$ 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 $(8,0)$.
$$
\begin{aligned}
& \therefore \quad 0=8 m \pm 2 \sqrt{1+m^2} \\
& \Rightarrow \quad-4 m= \pm \sqrt{1+m^2} \\
& \Rightarrow \quad 16 m^2=1+m^2 \\
& \Rightarrow \quad 15 m^2=1 \\
& \Rightarrow \quad m^2=\frac{1}{15} \\
& \Rightarrow \quad m= \pm \frac{1}{\sqrt{15}} \\
&
\end{aligned}
$$
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