The equation of the ellipse whose vertices are at (± 5, 0) and foci at (± 4, 0) is

Question

Question

Answer 1

Concept:

Equation of ellipse: \(\rm\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Eccentricity (e) = \(\rm\sqrt{1-\frac{b^2 }{a^2}}\)

Where, vertices = (± a, 0) and focus = (± ae, 0)

Calculation:

Here, vertices of ellipse (± 5, 0) and foci (±4, 0)

So, a = ±5 ⇒ \(a^2=25\) and

ae = 4 ⇒ e = 4/5

Now, 4/5 = \(\rm\sqrt{1-\frac{b^2 }{5^2}}\)

\(⇒ \rm\frac{16}{25}=\rm\frac{25-b^2}{25}\\⇒ 16=25-b^2 \\⇒ b^2=9 \)

∴ Equation of ellipse = \(\rm \frac {x^2}{25} + \frac {y^2}{9} = 1\)

Hence, option (1) is correct.