Question

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 meets 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 1

Ans: (A)
Hint: \(P(3 \cos \theta, 2 \sin \theta)\)
Equation of line // to \(y\) axis is \(x=3 \cos \theta\)
It meets circle at \(\theta\)
\(\therefore Q(3 \cos \theta, 3 \sin \theta)\)
\(\because \mathrm{PR}: \mathrm{RQ}=1: 2\)
\(\therefore R\left(3 \cos \theta, \frac{7 \sin \theta}{3}\right)\)
\(\Rightarrow \frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1\)

Answer 2

cheapest cialis online <a href="https://ordergnonline.com/">cialis dosage 40 mg</a> best ed pills online