Let the eccentricity of an ellipse x²/a² + y²/b² = 1, a > b, be 1/4. If this ellipse passes through the point ( - 4√(2/5) ,3 ), then a² + b² is equal to :

Solution

Maths sir · 4 steps

Use the eccentricity to write b² in terms of a², then the point gives you a single unknown.

1. Link b to a. For an ellipse with a > b, b² = a²(1 - e²). With e = 1/4, e² = 1/16, so b² = 15/16a².
2. Square the coordinates. At the given point, x² = 16 × 2/5 = 32/5 and y² = 9. Put these into the equation of the ellipse.

   32/5a² + (9 × 16)/15a² = 32/5a² + 48/5a² = 80/5a² = 16/a² = 1

3. Find both squares. So a² = 16, and then b² = 15/16 × 16 = 15.
4. So. a² + b² = 16 + 15 = 31, which is option (b).

The idea to keep. Eccentricity is a relation between a and b. Use b² = a²(1 - e²) to leave one unknown, and then a single point on the ellipse fixes it.

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