Question

The equation of the tangent to the parabola $y^2=8x$ is $y=x+2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

• A. (0 , 2)
• B. (2 , 4)
• C. (-2 , 0)
• D. (-1 , 1)

Answer 1

The correct option is C (-2 , 0)

$\begin{array}{c}\text{Given,}\\ y^2=8x\\ \text{Equation of the tangent is}y=x+2\\ \text{Slope of the tangent is,}m=1\\ \text{When}2\text{lines are perpendicular to each other, then}\\ \text{product of their slopes is}-1.\\ m_1=-1\\ \text{let other tangent be}\\ y=m_{1}x+c=-x+c\end{array}$

$\begin{array}{c}\text{A line touching the parabola is said to be a tangent to the parabola provided it satisfies}\\ \text{certain conditions.}\\ \text{If we have a line y}=m_{1}x+\text{c touching a parabola}{y}^2=4ax\\ \text{then we know, c}=\frac{a}{m_1}\\ y^2=8x\text{and}y=-x+c\\ \text{4a}=8\end{array}$

$\begin{array}{c}\text{4a}=8\\ a=2\\ c=\frac{a}{m_1}=\frac{2}{-1}=-2\\ \text{hence other tangent will be}y=-x-2\\ \text{Point of intersection of both tangents :}\end{array}$

$\begin{array}{c}\text{Point of intersection of both tangents :}\\ y=x+2\dots .\left(1\right)\\ y=-x-2\dots \left(2\right)\\ \text{from}\left(1\right)\text{and}\left(2\right),\text{we get}\end{array}$

$\begin{array}{c}x+2=-x-2\\ 2x=-4\\ x=-2\\ y=x+2=-2+2=0\\ \text{hence point is}(-2,0)\end{array}$