JEE Main 2021 · 16 March · evening · Single correct · +4 / −1
If the points of intersections of the ellipse x²/16 + y²/b² = 1 and the
circle x² + y² = 4b, b > 4 lie on the curve y² = 3x², then b is equal to :
- (a) 12
- (b) 10
- (c) 6
- (d) 5
Answer and solution
Answer (a) 12
Solution
Maths sir · 4 steps
Use y² = 3x² in the circle to get the point, then make that point satisfy the ellipse.
- Use the circle. The intersection points lie on y² = 3x², so put this into x² + y² = 4b. That gives 4x² = 4b, so x² = b and y² = 3b.
- Use the ellipse. The same points also lie on the ellipse, so put x² = b and y² = 3b into x²/16 + y²/b² = 1. This gives b/16 + 3b/b² = 1, which is b/16 + 3/b = 1.
b² - 16b + 48 = 0 ⇒ (b - 12)(b - 4) = 0
- Apply b > 4. The roots are b = 12 and b = 4. The condition b > 4 rules out 4. Check: with b = 12 we get x² = 12 and y² = 36, and 12/16 + 36/144 = 3/4 + 1/4 = 1.
- So. b = 12, which is option (a).
The idea to keep. When intersection points also lie on a third curve, use the simplest curve first to get x² and y². Then make those values satisfy the remaining equation.
Answer from the NTA final answer key. Solution by PrepWiser.