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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

If the point of intersections of the ellipse \(\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1\) and the circle \(x ^{2}+ y ^{2}=4 b , b > 4\) lie on the curve \(y^{2}=3 x^{2},\) then \(b\) is equal to:

  1. A \(12\)
  2. B \(5\)
  3. C \(6\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

\(y^{2}=3 x^{2}\) And \(x^{2}+y^{2}=4 b\) Solve both we get So \(x^{2}=b\) \(\frac{x^{2}}{16}+\frac{3 x^{2}}{b^{2}}=1\) \(\frac{b}{16}+\frac{3}{b}=1\) \(b^{2}-16 b+48=0\) \((b-12)(b-4)=0\) \(b=12, b > 4\)
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