A straight line parallel to the line $2x-y+5=0$ is also a tangent to the curve $y^2=4x+5$. Then, the point of contact is

The correct option is C $(-1,1)$
Given curve: $y^2=4x+5$

Equation of line parallel to $2x-y+5=0$ is $2x-y+k=0.$ where, $k$ is a constant.

Slope of the line $=2$ ...... (1)

Slope of tangent to the curve is

$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}$ ......(2)

From (1) and (2), we get

$\frac{2}{y}=2$

$\Rightarrow y=1$

Putting value $y=1$ in equation of curve, we get

$1^2=4x+5\Rightarrow 4x=-4$

$\Rightarrow x=-1$

Point of contact$=(-1,1)$

Hence, B is the correct option.