Question

The equation of the normal to the parabola $x^2=8y$ at $x=4$ is

• A. $x + 2y = 0$
• B. $x+y = 2$
• C. $x - 2y = 0$
• D. $x+y = 6$

Answer 1

Answer: (D)

$x+y = 6$

Answer 2

$x^{2}=8y\,...\left(i\right)$ When, $x = 4,$ then $y = 2$ Now $\frac{dy}{dx}=\frac{2x}{8}=\frac{x}{4}, \frac{dy}{dx}]_{x=4}=4$ Slope of normal $=-\frac{1}{\frac{dy}{dx}}=-1$ Euqation of normal at $x = 4$ is $y - 2 = - 1 \left(x - 4\right)$ $\Rightarrow y = -x + 4 + 2 = -x + 6$ $\Rightarrow x+y=6$