Question

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

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

Answer 1

Answer: (B)

$(-2,0)$

Given line is $y=x+2$

Slope of this line is $1$

Then, slope of the line perpendicular to the given line will be $-1$

Let $(x_1,y_1)$ be a point on $y=x+2$

$\therefore y_1=x_1+2$

Equation of the line perpendicular to the given line through $(x_1,y_1)$ is

$y-(x_1+2)=-(x-x_1)\Rightarrow y=-x+2(x_1+1)$

If this line is a tangent to $y^2=8x,$

$c=\frac{a}{m}$ gives

$2(x_1+1)=\frac{2}{-1}\Rightarrow x_1+1=-1\Rightarrow x_1=-2$

Hence $y_1=0$

$\therefore$ The required point is $(-2,0)$.