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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Ellipse

NDA 2018 · Mathematics · Ellipse

What is the equation of the ellipse whose vertices are ( ± 5,0) and...

What is the equation of the ellipse whose vertices are (±5,0)( \pm 5,0) and foci are at (±4,0)( \pm 4,0) ?

Options

  1. Ax225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1
  2. Bx216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1
  3. Cx225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1
  4. Dx29+y225=1\frac{x^2}{9}+\frac{y^2}{25}=1

Correct answer

A. x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1

Step-by-step solution

Given, vertices, (5,0)and(−5,0)(5,0) and (-5,0) foci (4,0)and(−4,0)(4,0) and (-4,0)
i.e., ae=4⇒e=4a=45a \mathrm{e}=4\Rightarrow \mathrm{e}=\frac{4}{\mathrm{a}}=\frac{4}{5} we know, b2=a2(1−e2)\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=52(1−(45)2)=25(25−1625)=9∴a2=25,b2=9\begin{aligned} & =5^2\left(1-\left(\frac{4}{5}\right)^2\right) \\ & =25\left(\frac{25-16}{25}\right)=9 \\ & \therefore \mathrm{a}^2=25, \mathrm{ b}^2=9 \end{aligned}$EquationofellipseisEquation of ellipse is\frac{x^2}{25}+\frac{y^2}{9}=1$
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