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

Let \(y^2=12 x\) be the parabola and \(S\) be its focus. Let PQ be a focal chord of the parabola such that \((\mathrm{SP})(\mathrm{SQ})=\frac{147}{4}\). Let C be the circle described taking PQ as a diameter. If the equation of a circle \(C\) is \(64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta\), then \(\beta-\alpha\) is equal to ________.

  1. A 1300
  2. B 1320
  3. C 1328
  4. D 1330
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Answer & Solution

Correct Answer

(C) 1328

Step-by-step Solution

Detailed explanation

\(\mathrm{y}^2=12 \mathrm{x} \quad \mathrm{a}=3 \quad \mathrm{SP} \times \mathrm{SQ}=\frac{147}{4}\) Let \(\mathrm{P}\left(3 \mathrm{t}^2, 6 \mathrm{t}\right)\) and \(\mathrm{t}_1 \mathrm{t}_2=-1\) (ends of focal chord) So, \(Q\left(\frac{3}{t^2}, \frac{-6}{t}\right)\)…
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