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If the focal distance of a point $\mathrm{P}\left(2, \mathrm{y}_1\right)$ on the parabola $\mathrm{y}^2=\mathrm{kx}$ is 3 , then the equation of the tangent drawn at $\mathrm{P}$ to the given parabola is
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Verified Answer
The correct answer is:
$x \pm \sqrt{2} y+2=0$
Since $P\left(2, y_1\right)$ lies on $y^2=k x$
$$
\begin{aligned}
& \Rightarrow y_1{ }^2=2 k \Rightarrow y_1=\sqrt{2 k} \\
& \therefore P\left(2, y_1\right) \equiv P(2, \sqrt{2 k})
\end{aligned}
$$
Focus of $y^2=k x$ will be $\left(\frac{k}{4}, 0\right)$
$$
\begin{aligned}
& \text { Given } \sqrt{\left(2-\frac{k}{4}\right)^2+(\sqrt{2 k}-0)^2}=3 \\
& \Rightarrow k^2+16 k-80=0 \Rightarrow(k+20)(k-4)=0 \\
& \Rightarrow k=4, k=-20(\text { Rejected }) \\
& \therefore P\left(2, y_1\right) \equiv P(2, \sqrt{2 \times 4}) \equiv P(2, \pm \sqrt{2})
\end{aligned}
$$
and $y^2=k x=4 x$
Hence equation of tangent at $P(2, \pm 2 \sqrt{2})$ on $y^2=4 x$ is
$$
\begin{aligned}
& \Rightarrow y( \pm 2 \sqrt{2})=2(x+2) \\
& \Rightarrow x \pm \sqrt{2} y+2=0
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow y_1{ }^2=2 k \Rightarrow y_1=\sqrt{2 k} \\
& \therefore P\left(2, y_1\right) \equiv P(2, \sqrt{2 k})
\end{aligned}
$$
Focus of $y^2=k x$ will be $\left(\frac{k}{4}, 0\right)$
$$
\begin{aligned}
& \text { Given } \sqrt{\left(2-\frac{k}{4}\right)^2+(\sqrt{2 k}-0)^2}=3 \\
& \Rightarrow k^2+16 k-80=0 \Rightarrow(k+20)(k-4)=0 \\
& \Rightarrow k=4, k=-20(\text { Rejected }) \\
& \therefore P\left(2, y_1\right) \equiv P(2, \sqrt{2 \times 4}) \equiv P(2, \pm \sqrt{2})
\end{aligned}
$$
and $y^2=k x=4 x$
Hence equation of tangent at $P(2, \pm 2 \sqrt{2})$ on $y^2=4 x$ is
$$
\begin{aligned}
& \Rightarrow y( \pm 2 \sqrt{2})=2(x+2) \\
& \Rightarrow x \pm \sqrt{2} y+2=0
\end{aligned}
$$
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