Question

The foci of an ellipse are $\left(-2,4\right)$ and $\left(2,1\right)$ . The point $\left(1,\frac{23}{6}\right)$ is an extremity on the minor axis. The value of the eccentricity is $\frac{a}{\sqrt{b}}$ then the value of $a+b$ is

Answer 1

Answer: 16

Given: Foci of the ellipse are $A\left(-2,4\right)$ and $B\left(2,1\right)$ .
Center is the midpoint of the line joining foci, $AB$ i.e., $C\left(0,\frac{5}{2}\right)$ .
$b=$ Distance of center from the extremity of the minor axis $D\left(1,\frac{23}{6}\right)$ .
Now using distance formula,
$\therefore b=\sqrt{1^2+{\left(\frac{23}{6}-\frac{5}{2}\right)}^2}=\frac{5}{3}$
We know distance between foci $=2ae$ i.e.
$AB=2ae=\sqrt{{\left(2+2\right)}^2+{\left(4-1\right)}^2}$
Now, $b^2=a^2-a^2{e}^2$
$\Rightarrow a^2=b^2+a^2{e}^2=\frac{325}{36}$
$\Rightarrow e^2=1-\frac{b^2}{a^2}=\frac{9}{13}$
$\Rightarrow e=\frac{3}{\sqrt{13}}$