Q. 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
NTA AbhyasNTA Abhyas 2022
Solution:
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} \, $
$\Rightarrow e=\frac{3}{\sqrt{13}}$
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} \, $
$\Rightarrow e=\frac{3}{\sqrt{13}}$