Question

$S$ and $S^{{}^{\prime }}$ are the foci of the ellipse $25x^2+16y^2=1600$. Then the area of the triangle formed by the foci $S$ and $S^{{}^{\prime }}$ with the point $(4\sqrt{3},5)$ is

• A. $24\sqrt{3}$
• B. $25\sqrt{3}$
• C. $40\sqrt{3}$
• D. $30\sqrt{3}$

Answer 1

The correct option is A $24\sqrt{3}$

Ellipse is

$\frac{x^2}{64}+\frac{y^2}{100}=1$
Let $A,B$ be the foci and point $P=(4\sqrt{3},5)$

We need to find area of triangle $ABP$.
Area of a triangle is $\frac{1}{2}\ast base\ast height=\frac{1}{2}\ast AB\ast height$
Height is nothing but the $X$ co ordinate of point $P=4\sqrt{3}$ as the major axis is along $Y$ axis
$AB=2be=2\ast 10\ast \sqrt{(1-\frac{64}{100})}=2\ast 10\ast \sqrt{\frac{36}{100}}=2\ast 10\ast \frac{6}{10}=12$
Therefore, area$=\frac{1}{2}\ast 4\sqrt{3}\ast 12=24\sqrt{3}$