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If $5 x-2 y+k=0$ is a tangent to the parabola $y^2=6 x$, then their point of contact is
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Verified Answer
The correct answer is:
$\left(\frac{6}{25}, \frac{6}{5}\right)$
Given curve
$$
y^2=6 x
$$
difference w.r.t ' $x$ '
$$
\Rightarrow \quad 2 y \frac{d y}{d x}=6
$$
Given equation of tangent is $5 x-2 y+k=0$
$$
\begin{aligned}
\quad \text { Slope } & =\frac{5}{2} \\
\therefore \quad \frac{3}{y} & =\frac{5}{2} \Rightarrow y=\frac{6}{5}
\end{aligned}
$$
Substitute value of $y$ in $y^2=6 x$
$$
\begin{aligned}
\left(\frac{6}{5}\right)^2 & =6 x \Rightarrow \frac{36}{25}=6 x \\
\Rightarrow \quad x & =\frac{6}{25}
\end{aligned}
$$
Hence, point of contact is $\left(\frac{6}{25}, \frac{6}{5}\right)$
$$
y^2=6 x
$$
difference w.r.t ' $x$ '
$$
\Rightarrow \quad 2 y \frac{d y}{d x}=6
$$
Given equation of tangent is $5 x-2 y+k=0$
$$
\begin{aligned}
\quad \text { Slope } & =\frac{5}{2} \\
\therefore \quad \frac{3}{y} & =\frac{5}{2} \Rightarrow y=\frac{6}{5}
\end{aligned}
$$
Substitute value of $y$ in $y^2=6 x$
$$
\begin{aligned}
\left(\frac{6}{5}\right)^2 & =6 x \Rightarrow \frac{36}{25}=6 x \\
\Rightarrow \quad x & =\frac{6}{25}
\end{aligned}
$$
Hence, point of contact is $\left(\frac{6}{25}, \frac{6}{5}\right)$
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