The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latusrectum is $\frac{15}{2}$ units, is

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Standard form of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ a > b $\because$ Distance between foci $=2 a e$ 2 = 2ae

$\Rightarrow \quad a e=1$...(ii) $\begin{aligned} & \text { and length of latusrectum }=\frac{2 b^2}{a} \\ & \qquad \begin{aligned} \frac{15}{2} & =\frac{2 b^2}{a} \Rightarrow b^2=\frac{15 a}{4} \\ \because \quad b^2 & =a^2\left(-e^2+1\right)=-a^2 e^2+a^2 \\ \frac{15 a}{4} & =-1+a^2\end{aligned}\end{aligned}$ $\begin{aligned} & \Rightarrow \quad 4 a^2-15 a-4=0 \\ & 4 a^2-16 a+a-4=0 \\ & 4 a(a-4)+(a-4)=0 \\ &(4 a+1)(a-4)=0 \\ & \Rightarrow \quad a=-1 / 4 \text { and } 4\end{aligned}$ When, a = + 4 $ \Rightarrow \quad b^2=\frac{15 a}{4}=\frac{15 \times 4}{4} \Rightarrow b=\sqrt{15} $ $\therefore$ Equation of ellipse, $\frac{x^2}{16}+\frac{y^2}{15}=1$ $ \Rightarrow \quad 15 x^2+16 y^2=240 $

The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the…

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