Ellipse 1 Question 11
12. Let $P$ be a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,0<b<a$. Let the line parallel to $Y$-axis passing through $P$ meet the circle $x^{2}+y^{2}=a^{2}$ at the point $Q$ such that $P$ and $Q$ are on the same side of $X$-axis. For two positive real numbers $r$ and $s$, find the locus of the point $R$ on $P Q$ such that $P R: R Q=r: s$ as $P$ varies over the ellipse.
$(2001,4 M)$
Passage Type Questions
Passage
Let $F _1\left(x _1, 0\right)$ and $F _2\left(x _2, 0\right)$, for $x _1<0$ and $x _2>0$, be the foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$. Suppose a parabola having vertex at the origin and focus at $F _2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
(2016 Adv.)
Show Answer
Answer:
Correct Answer: 12. (a)
Solution:
- Given, $\frac{P R}{R Q}=\frac{r}{s}$
$$ \begin{array}{rlrl} \Rightarrow & & \frac{\alpha-b \sin \theta}{a \sin \theta-\alpha} & =\frac{r}{s} \\ \Rightarrow & \alpha s-b \sin \theta \cdot s & =r a \sin \theta-\alpha r \\ \Rightarrow & \alpha s+\alpha r & =r a \sin \theta+b \sin \theta \cdot s \\ \Rightarrow & \alpha(s+r) & =\sin \theta(r a+b s) \\ \Rightarrow & \alpha & \alpha & =\frac{\sin \theta(r a+b s)}{r+s} \end{array} $$
Let the coordinates of $R$ be $(h, k)$.
$\therefore \quad h=a \cos \theta \Rightarrow \cos \theta=\frac{h}{a}$
and
$$ k=\alpha=\frac{(a r+b s) \sin \theta}{r+s} $$
$$ \Rightarrow \quad \sin \theta=\frac{k(r+s)}{a r+b s} $$
On squaring and adding Eqs. (i) and (ii), we get
$$ \begin{aligned} \sin ^{2} \theta+\cos ^{2} \theta & =\frac{h^{2}}{a^{2}}+\frac{k^{2}(r+s)^{2}}{(a r+b s)^{2}} \\ \Rightarrow \quad 1 & =\frac{h^{2}}{a^{2}}+\frac{k^{2}(r+s)^{2}}{(a r+b s)^{2}} \end{aligned} $$
Hence, locus of $R$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}(r+s)^{2}}{(a r+b s)^{2}}=1$.
