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

If the points of intersection of two distinct conics \(x^2+y^2=4 b\) and \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) lie on the curve \(y^2=3 x^2\), then \(3 \sqrt{3}\) times the area of the rectangle formed by the intersection points is ...........

  1. A \(432\)
  2. B \(456\)
  3. C \(123\)
  4. D \(789\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(432\)

Step-by-step Solution

Detailed explanation

Putting \(y^2=3 x^2\) in both the conics We get \(x^2=b\) and \(\frac{b}{16}+\frac{3}{b}=1\) \(\Rightarrow b=4,12\) (b \(=4\) is rejected because curves coincide) \(\therefore \mathrm{b}=12\) Hence points of intersection are…
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