Question

The foci of an ellipse are located at the points $(2,4)$ and $(2,-2)$. The points $(4,2)$ lies on the ellipse. If $a$ and $b$ represent the lengths of the semi-major and semi-minor axes respectively, then the value of $(ab{)}^2$ is equal to

• A. $68+22\sqrt{10}$
• B. $6+22\sqrt{10}$
• C. $26+10\sqrt{10}$
• D. $6+10\sqrt{10}$

Answer 1

Answer: (C)

The distance between the foci is 6 , so $c=3$.
The sum of the distances from $(4,2)$ to each of the foci is the major axis length,
so $2a=\sqrt{(4-2{)}^2+(2-4{)}^2}+\sqrt{(4-2{)}^2+(2+2{)}^2}=\sqrt{4+4}+\sqrt{4+16}=\sqrt{8}+\sqrt{20}$
$=2\sqrt{2}+2\sqrt{5}\Rightarrow a=\sqrt{2}+\sqrt{5}$
Also, for an ellipse, $b^2=a^2-c^2=(\sqrt{2}+\sqrt{5}{)}^2-3^2=7+2\sqrt{10}=-2+2\sqrt{10}$.
Thus, we have $(ab{)}^2=(7+2\sqrt{10})(-2+2\sqrt{10})$
$=-14+14\sqrt{10}-4\sqrt{10}+40=26+10\sqrt{10}$