The length of the latus rectum of an ellipse is 18/5 and eccentricity is 4/5, then equation of the ellipse is…

Correct answer: (B) (x^2)/25+(y^2)/9=1. Given, (2b^2)/a=18/5⇒(b^2)/a=9/5 ⇒b^2=9/5a …(i) and, e=√(1-b^2a^2)=4/5⇒1-(b^2)/(a^2)=16/25 ⇒(b^2)/(a^2)=1-16/25=9/25 ⇒9/5aa^2=9/25⇒9/5a=9/25 ⇒a=5 ∴b^2=9/55=9 [using Eq. (i)] ∴ Required equation of ellipse, (x^2)/25+(y^2)/9=1

Q53 / 150 · MHT CET 2019 (Shift 1) · Mathematics

The length of the latus rectum of an ellipse is $\frac{18}{5}$ and eccentricity is $\frac{4}{5},$ then equation of the ellipse is…

A
$\frac{x^{2}}{25} + \frac{y^{2}}{8} = 1$
Correct answer
B
$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$
C
$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
D
$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

Solution

Given,

\frac{2 b^{2}}{a} = \frac{18}{5} \Rightarrow \frac{b^{2}}{a} = \frac{9}{5}

\Rightarrow b^{2} = \frac{9}{5} a

…(i)
and,

e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{4}{5} \Rightarrow 1 - \frac{b^{2}}{a^{2}} = \frac{16}{25}

\Rightarrow \frac{b^{2}}{a^{2}} = 1 - \frac{16}{25} = \frac{9}{25}

\Rightarrow \frac{(\frac{9}{5}) a}{a^{2}} = \frac{9}{25} \Rightarrow \frac{9}{5 a} = \frac{9}{25}

\Rightarrow a = 5

∴ b^{2} = \frac{9}{5} (5) = 9

[using Eq. (i)]

∴

Required equation of ellipse,

\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1

Hence, the correct answer is option: 'B'.

The length of the latus rectum of an ellipse is 185 and eccentricity is 45, then equation of the ellipse is…

Solution

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The length of the latus rectum of an ellipse is $\frac{18}{5}$ and eccentricity is $\frac{4}{5},$ then equation of the ellipse is…