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

Let the tangent and normal at the point \((3 \sqrt{3}, 1)\) on the ellipse \(\frac{x^2}{36}+\frac{y^2}{4}=1\) meet the \(y\)-axis at the points \(A\) and \(B\) respectively. Let the circle \(C\) be drawn taking \(A B\) as a diameter and the line \(x =2 \sqrt{5}\) intersect \(C\) at the points \(P\) and \(Q\). If the tangents at the points \(P\) and \(Q\) on the circle intersect at the point \((\alpha, \beta)\), then \(\alpha^2-\beta^2\) is equal to

  1. A \(\frac{314}{5}\)
  2. B \(\frac{304}{5}\)
  3. C \(60\)
  4. D \(61\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{304}{5}\)

Step-by-step Solution

Detailed explanation

Given ellipse \(\frac{x^2}{36}+\frac{y^2}{4}=1\) \(\frac{x}{4 \sqrt{3}}+\frac{y}{4}=1\) \(y=4\) \(\frac{x}{4}-\frac{4}{4 \sqrt{3}}=\frac{2}{\sqrt{3}}\) \(y=-8\) \(x^2+y^2+4 y-32=0\) \(h x+k y+2(y+k)-32=0\) \(k=-2\) \(h x+2 k-32=0\) \(h x=36\) \(\alpha=h=\frac{36}{2 \sqrt{5}}\)…
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