Question:
The equation of the normal to the parabola $x^2=8y$ at $x=4$ is
The equation of the normal to the parabola $x^2=8y$ at $x=4$ is
Updated On: Jul 14, 2022
- $x + 2y = 0$
- $x+y = 2$
- $x - 2y = 0$
- $x+y = 6$
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The Correct Option is D
Solution and Explanation
$x^{2}=8y\,...\left(i\right)$
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
$y - 2 = - 1 \left(x - 4\right)$
$\Rightarrow y = -x + 4 + 2 = -x + 6$
$\Rightarrow x+y=6$
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
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