WBJEE · Maths · Ellipse
Let \(P\) be a point on the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) and the line through \(P\) parallel to the \(Y\) -axis mects the circle \(x^{2}+y^{2}=9\) at \(Q\) where \(P, Q\) are on the same side of the \(X\) -axis. If \(R\) is a point on \(P Q\) such that \(\frac{P R}{R Q}=\frac{1}{2},\) then the locus of \(R\) is
- A \(\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1\)
- B \(\frac{x^{2}}{49}+\frac{y^{2}}{9}=1\)
- C \(\frac{x^{2}}{9}+\frac{y^{2}}{49}=1\)
- D \(\frac{9 x^{2}}{49}+\frac{y^{2}}{9}=1\)
Answer & Solution
Correct Answer
(A) \(\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1\)
Step-by-step Solution
Detailed explanation
Since, point \(P\) on the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) \(\therefore\) \(P(3 \cos \theta, 2 \sin \theta)\) Now, equation of line parallel of \(Y\) -axis is \[ x=3 \cos \theta \] and above line meets circle at \(Q\) \(\therefore\)…
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