Let the line y = mx and the ellipse 2x² + y² = 1 intersect at a ponit P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at ( - 1/3√2,0 ) and (0, β), then β is equal to :

Solution

Maths sir · 5 steps

Write the normal at a general point P, then let the given x-intercept tell you exactly where P is.

1. Slope of the normal. Let P = (x₁, y₁). Differentiating 2x² + y² = 1 gives 4x + 2y y' = 0, so the tangent slope at P is -2x₁/y₁. The normal slope is therefore y₁/2x₁.

   y - y₁ = y₁/2x₁(x - x₁)

2. Use the x-intercept. Put y = 0: -y₁ = y₁/2x₁(x - x₁), so x - x₁ = -2x₁ and x = -x₁. Matching with -1/3√2 gives x₁ = 1/3√2.
3. Find P. From the ellipse, y₁² = 1 - 2x₁² = 1 - 2/18 = 8/9. P is in the first quadrant, so y₁ = 2√2/3. The line y = mx only fixes the quadrant here; you never need m.
4. Use the y-intercept. Put x = 0 in the normal: y = y₁ - y₁/2x₁ × x₁ = y₁/2.
5. So. β = 1/2 × 2√2/3 = √2/3, which is option (a).

The idea to keep. When a normal's intercepts are given, find both intercepts in terms of (x₁, y₁) first. One known intercept pins down P, and the ellipse equation then gives the other coordinate.

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