The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x=-4, find the equation of the normal to the ellipse at the point (1,3/2).

D. 4x − 2y = 1. Step 1: Directrix x=-a/e=-4 with e=1/2 gives a=2. Step 2: b^2=a^2(1-e^2)=4× 3/4=3, so the ellipse is x^24+y^23=1. Check the point: 1/4+9/4/3=1/4+3/4=1, so (1,3/2) lies on it. Step 3: Normal at (x_1,y_1): a^2xx_1-b^2yy_1=a^2-b^2 Step 4: 4x/1-3y/3/2=4-3 4x-2y=1 Trap: 4x+2y=7 is not the normal — it does not even pass through (1,3/2) with the right slope ( -2 instead of 2).

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The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , find the equation of the normal to the ellipse at the point $(1, \frac{3}{2})$ .

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Answer: Option D — 4x − 2y = 1

Detailed Solution

Directrix $x = - \frac{a}{e} = - 4$ with $e = \frac{1}{2}$ gives $a = 2$ .

$b^{2} = a^{2} (1 - e^{2}) = 4 \times \frac{3}{4} = 3$ , so the ellipse is $\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$ .

Check the point: $\frac{1}{4} + \frac{9/4}{3} = \frac{1}{4} + \frac{3}{4} = 1$ , so $(1, \frac{3}{2})$ lies on it.

Normal at $(x_{1}, y_{1})$ : $\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}$

$\frac{4 x}{1} - \frac{3 y}{3/2} = 4 - 3 \Rightarrow 4 x - 2 y = 1$

⚠ Common trap

4 x + 2 y = 7

is not the normal — it does not even pass through

(1, \frac{3}{2})

with the right slope (

- 2

instead of

2

Quick tip

Normal to ellipse at $(x_{1}, y_{1})$ : $\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}$

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Ellipse — Preparation

This question tests your understanding of Ellipse, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

An ellipse is the set of points whose distances from two fixed points (the foci) add up to a constant. Its standard form is x^2/a^2 + y^2/b^2 = 1, where a > b, and its eccentricity e satisfies b^2 = a^2(1 - e^2), with 0 < e < 1. The foci are at (ae, 0) and (-ae, 0) and the directrices are x = a/e and x = -a/e. Exam questions test finding the foci, eccentricity, length of the major and minor axes, latus rectum (2b^2/a), focal distances, and equations of tangents and normals. A useful fact is that the sum of the focal distances of any point on the ellipse equals the length of the major axis, 2a. Always check whether the major axis lies along the x-axis or the y-axis before applying formulas. Practise the ellipse questions below, each with a step-by-step solution.

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The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , find the equation of the normal to the ellipse at the point $(1, \frac{3}{2})$ .

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The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directrices is $x = - 4$ , find the equation of the normal to the ellipse at the point $(1, \frac{3}{2})$ .