Question

For an ellipse with axes are coordinate axes, $A$ and $L$ are the ends of major axis and latusrectum respectively. Area of $△OAL=8sq.$ units and, $e=\frac{1}{\sqrt{2}}$ , then equation of the ellipse is:

• A. $\frac{x^2}{16}+\frac{y^2}{8}=1$
• B. $\frac{x^2}{32}+\frac{y^2}{16}=1$
• C. $\frac{x^2}{64}+\frac{y^2}{32}=1$
• D. $\frac{x^2}{8}+\frac{y^2}{4}=1$

Answer 1

The correct option is B $\frac{x^2}{32}+\frac{y^2}{16}=1$

The area of the triangle with vertices $O(0,0)A(0,b)L(ae,\frac{b^2}{a})$ is:

$A=\frac{1}{2}\times base\times height$

$\Rightarrow A=\frac{1}{2}\times b\times ae$

$\Rightarrow \frac{abe}{2}=8$

$\Rightarrow ab=16\sqrt{2}$ -----------(1)

And, $b^2=a^2-{\left(ae\right)}^2$

$\Rightarrow b^2=a^2-\frac{a^2}{2}$

$\Rightarrow b^2=\frac{a^2}{2}$

By using equation 1 we get:

$\frac{a^2}{\sqrt{2}}=16\sqrt{2}$

$\Rightarrow a=4\sqrt{2}$ and $b=4$

$\therefore$ Equation of the ellipse is:

$\frac{x^2}{32}+\frac{y^2}{16}=1$