Question

An ellipse passing through the point $(2\sqrt{13},4)$ has its foci at $(-4,1)$ and $(4,1)$, then its eccentricity is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Let foci $A=(-4,1)B=(4,1)$ and point be $P=(2\sqrt{13},4)$
Distance between foci is $AB=2ae=\sqrt{(4+4{)}^2}=8$
$AP+BP=2a$
$AP=\sqrt{(4+2\sqrt{13}{)}^2+(4-1{)}^2}=\sqrt{16+16\sqrt{13}+52+9}$
$=\sqrt{77+16\sqrt{13}}=\sqrt{64+13+2\times 8\times \sqrt{13}}$
$=\sqrt{(8+\sqrt{13}{)}^2}=8+\sqrt{13}$
Similarly $BP=\sqrt{(4-2\sqrt{13}{)}^2+(4-1{)}^2}=\sqrt{77-16\sqrt{13}}=8-\sqrt{13}$
$AP+BP=8+\sqrt{13}+8-\sqrt{13}=16$
Therefore $e=\frac{AB}{AP+BP}=\frac{8}{16}=\frac{1}{2}$