JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q80

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MathematicsSingle correct+4 / -1Question 80 of 90

The eccentricity of an ellipse whose centre is at the origin is $\dfrac12$ . If one of its directrices is $x=-4$ , then the equation of the normal to it at $\left(1,\dfrac32\right)$ is :

4x-2y=1

4x+2y=7

x+2y=4

2y-x=2

Answer and worked solution

Official answer: A

$\frac ae=4$ gives $a=2$ , and $b^2=a^2(1-e^2)=3$ . The normal to $\frac{x^2}4+\frac{y^2}3=1$ at $\left(1,\frac32\right)$ is $\dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2$ :
$4x-2y=1.$

Topics:Ellipse Tangents and normals to conics

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 80