Exams › JEE Advanced › Maths

If the curves x²/a + y²/4 = 1 and y³ = 16x intersect at right angles, then a possible value of a is

1. 4/3
2. 3/4
3. 1/2
4. 2

Correct answer: 4/3

Solution

Differentiating the ellipse gives slope -4x/(a y); differentiating y³ = 16x gives slope 16/(3y²). Orthogonality and the curve equations lead to a = 4/3.

Related JEE Advanced Maths questions

• What is the proportion between the height of the cone with the largest possible volume that can fit inside a sphere and the sphere's diameter?
• Identify the point on the curve 9y² = x³ where the normal to the curve intersects the coordinate axes at equal distances.
• The normal to the parametric curve defined by x = a(cos θ + sin θ) and y = a(sin θ − cos θ) at any given value of θ:
• How many critical points exist for the function y = f(x) within the interval x ∈ [0, 2], if the function satisfies d²y/dx² = 6x - 4 and has a local minimum value of 5 at x = 1?
• What is the absolute minimum value of y = f(x) on the interval x ∈ [0, 2], given that f(x) satisfies the equation d²y/dx² = 6x - 4 and has a local minimum value of 5 at x = 1?
• The function f(x) = 2 |x| + |x+2| - ||x+2|-2| - |2|x|| has a local minimum or a local maximum at x =

⚔️ Practice JEE Advanced Maths free + battle 1v1 →