Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. If P(1, β), β > 0 is a point on this ellipse, then the equation of the normal to it at P is :

Solution

Maths sir · 4 steps

Use the directrix and e to find a and b, locate P, then apply the standard normal formula.

1. Find a and b. For an ellipse centred at the origin, the directrix is x = a/e. So a/(1/2) = 4, giving a = 2 and a² = 4. Then b² = a²(1 - e²) = 4 × 3/4 = 3, so the ellipse is x²/4 + y²/3 = 1.
2. Locate P. Put x = 1: 1/4 + β²/3 = 1, so β² = 9/4. Since β > 0, β = 3/2 and P = (1, 3/2).

   (a² x)/x₁ - (b² y)/y₁ = a² - b²

3. Put in numbers. With x₁ = 1 and y₁ = 3/2: 4x/1 - 3y/(3/2) = 4 - 3, which is 4x - 2y = 1.
4. So. The normal at P is 4x - 2y = 1, option (d). Check: it passes through P since 4 - 3 = 1, and its slope 2 is perpendicular to the tangent slope -1/2.

The idea to keep. For x²/a² + y²/b² = 1, the normal at (x₁, y₁) is a²x/x₁ - b²y/y₁ = a² - b². Get a and b first from the directrix x = a/e.

Answer from the NTA final answer key. Solution by PrepWiser.