Search any question & find its solution
Question:
Answered & Verified by Expert
Consider any point $P$ on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ in the first quadrant. Let $\mathrm{r}$ and s represent its distances from $(4,0)$ and
$(-4,0)$ respectively, then $(\mathrm{r}+\mathrm{s})$ is equal to
Options:
$(-4,0)$ respectively, then $(\mathrm{r}+\mathrm{s})$ is equal to
Solution:
1587 Upvotes
Verified Answer
The correct answer is:
10 unit
$\frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$
Put $\mathrm{x}=3$
$\frac{9}{25}+\frac{y^{2}}{9}=1$
$y=\frac{12}{5}$
$\mathrm{P}=(3,12 / 5)$
$r=P O=\sqrt{(4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=13 / 5$
$\mathrm{S}=\mathrm{PO}^{\prime}=\sqrt{(-4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=37 / 5$
Put $\mathrm{x}=3$
$\frac{9}{25}+\frac{y^{2}}{9}=1$
$y=\frac{12}{5}$
$\mathrm{P}=(3,12 / 5)$
$r=P O=\sqrt{(4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=13 / 5$
$\mathrm{S}=\mathrm{PO}^{\prime}=\sqrt{(-4-3)^{2}+\left(0-\frac{12}{5}\right)^{2}}$
$=37 / 5$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.