Question

A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ intersects the major and minor axes in points A and B respectively. If C is the center of the ellipse then the area of the triangle ABC is:

• A. 12 sq.units
• B. 24 sq.units
• C. 36 sq.units
• D. 48 sq.units

Answer 1

The correct option is B 24 sq.units

Comparing the equation $\frac{x^2}{18}+\frac{y^2}{32}=1$ with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get,

$\Rightarrow$ $a^2=18$ and $b^2=32$

$y=mx\pm \sqrt{a^2{m}^2+b^2}$

$\Rightarrow$ $y=\frac{-4}{3}x\pm \sqrt{18\times \frac{16}{9}+32}$

$\Rightarrow$ $y=\frac{-4}{3}x\pm \sqrt{64}$

$\Rightarrow$ $y=\frac{-4}{3}x\pm 8$

$\Rightarrow$ $3y=-4x\pm 24$

$\Rightarrow$ $4x+3y=\pm 24$

Now, for x-axis $y=0$

$\Rightarrow$ $4x=24$

$\Rightarrow$ $x=6$

for y-axis $x=0$

$\Rightarrow$ $3y=24$

$\Rightarrow$ $y=8$

Now, area of $△ABC=\frac{1}{2}\times 6\times 8=24$