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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Ellipse

NDA 2020 · Mathematics · Ellipse

Let P(x, y) be any point on the ellipse 25 x^2+16 y^2=400 . If...

Let P(x,y)P(x, y) be any point on the ellipse 25x2+16y2=40025 x^2+16 y^2=400. If Q(0,3)Q(0,3) and R(0,−3)R(0,-3) are two points, then what is (PQ+PR)(P Q+P R) equal to?

Options

  1. A12
  2. B10
  3. C8
  4. D6

Correct answer

B. 10

Step-by-step solution

Equation of the Ellipse : 25x2+16y2=40025 x^2+16 y^2=400. x216+y225=1\frac{x^2}{16}+\frac{y^2}{25}=1 Here a2=16,b2=25a^2=16, b^2=25 Focci points =(0,±be)=(0,±b2−a2)=(0,±25−16)=(0,±3)\begin{aligned} \text{ Focci points } & =(0, \pm b e)=\left(0, \pm \sqrt{b^2-a^2}\right) \\ & =(0, \pm \sqrt{25-16})=(0, \pm 3) \end{aligned}$PointPointQ=(0,3),and, andR=(0,-3).Now,fromquestionpoint. Now, from question pointP(x, y)liesontheellipse,thensumofitsdistancefromtwofixedpointsisalwaysconstant,equaltolengthofitsmajoraxis.Thus,lies on the ellipse, then sum of its distance from two fixed points is always constant, equal to length of its major axis. Thus,(P Q+P R)=2 \times 5=10$.
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