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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Ellipse

NDA 2015 · Mathematics · Ellipse

Consider any point P on the ellipse x^2 25 + y^2 9 =1 in the first...

Consider any point P on the ellipse x225+y29=1\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1 in the first quadrant. Let rr and ss represent its distances from (4,0)(4,0) and (−4,0)(-4,0) respectively, then (r+s)(r+s) is equal to

Options

  1. A10 unit
  2. B9 unit
  3. C8 unit
  4. D6 unit

Correct answer

A. 10 unit

Step-by-step solution

x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1 Put x=3\mathrm{x}=3925+y29=1y=125P=(3,12/5)r=PO=(4−3)2+(0−125)2=13/5S=PO′=(−4−3)2+(0−125)2=37/5r+s=135+375=505=10 unit \begin{aligned} & \frac{9}{25}+\frac{y^2}{9}=1 \\ & \mathrm{y}=\frac{12}{5} \\ & \mathrm{P}=(3,12 / 5) \\ & \mathrm{r}=\mathrm{PO}=\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 \\ & r+s=\frac{13}{5}+\frac{37}{5}=\frac{50}{5}=10 \text{ unit } \end{aligned}$$
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