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JEE Main 2017

JEE Main 2017 Mathematics question on Conic Sections

The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x=−4x = -4, then the equation of the normal to it at (1,32)\left(1, \dfrac{3}{2}\right) is
  1. A.
    2y−x=22y - x = 2
  2. B.
    4x−2y=14x - 2y = 1
  3. C.
    4x+2y=74x + 2y = 7
  4. D.
    x+4y=2x + 4y = 2
Show answer and solution

Correct option: B

Solution

For an ellipse with centre at origin, eccentricity e = 1/2. The directrix corresponding to the left side is x = -a/e = -2a. Given x = -4, so 2a = 4, meaning a = 2.

Then b² = a²(1 - e²) = 4(1 - 1/4) = 3, so b = √3.

The ellipse is x²/4 + y²/3 = 1.

For the normal at point (x₁, y₁), the equation is:
(a²x)/x₁ - (b²y)/y₁ = a² - b²

At (1, 3/2): (4x)/1 - (3y)/(3/2) = 4 - 3
→ 4x - 2y = 1.

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