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A straight line parallel to the line $2 \mathrm{x}-\mathrm{y}+5=0$ is also a tangent to the curve $y^{2}=4 x+5$. Then the point of contact is
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The correct answer is:
$(-1,1)$
Given curve is $y^{2}=4 x+5$ on differentiating, we get
$2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}$
Given line is $2 x-y+5=0$
$\Rightarrow y=2 x+5$
slope of line is 2 . Therefore,
$\frac{2}{y}=2 \Rightarrow y=1$
put $\mathrm{y}=1$ in the equation of curve, we get
$\begin{array}{l}
1=4 x+5 \\
x=-1
\end{array}$
Hence, point of contact is $(-1,1)$
$2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}$
Given line is $2 x-y+5=0$
$\Rightarrow y=2 x+5$
slope of line is 2 . Therefore,
$\frac{2}{y}=2 \Rightarrow y=1$
put $\mathrm{y}=1$ in the equation of curve, we get
$\begin{array}{l}
1=4 x+5 \\
x=-1
\end{array}$
Hence, point of contact is $(-1,1)$
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