Question

The locus of the foot of perpendicular drawn from focus upon a variable tangent to the parabola $2x-y+1^2=\frac{8}{\sqrt{5}}x+2y+3$ is

• A. $3x+y+4=0$
• B. $2x-y+1=0$
• C. $x+2y+3=0$
• D. None of these

Answer 1

Answer: (C)

As we know the property of parabola i.e. the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex.
Given the equation of parabola ${\left(2x-y+1\right)}^2=\frac{8\left(x+2y+3\right)}{\sqrt{5}}$
From standard equation of parabola $y^2=4ax$
we can say that $\left|y\right|$ is distance of a point from axis of parabola and $\left|x\right|$ is distance from tangent at vertex.
Similarly, in the give equation we can say that $2x-y+1=0$ is axis of parabola and $x+2y+3=0$ is tangent at the vertex.