Question

Area of the triangle formed by the $x$ axis, the tangent and normal at $(3,2)$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{8}=1$ is

• A. $5$
• B. $\frac{13}{3}$
• C. $\frac{15}{2}$
• D. $\frac{9}{2}$

Answer 1

Answer: (B)

$\frac{13}{3}$

Given equation of ellipse $\frac{x^2}{18}+\frac{y^2}{8}=1$
$\frac{dy}{dx}=\frac{-4x}{9y}$
Slope of tangent to ellipse at $(3,2)$ is
$m=\frac{-2}{3}$
Equation of tangent to ellipse is
$y-2=-\frac{2}{3}(x-3)$
$\Rightarrow 2x+3y=12$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=6$
So, tangent intersects x-axis at $(6,0)$
Equation of normal to ellipse is
$y-2=\frac{3}{2}(x-3)$
$\Rightarrow 3x-2y=5$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=\frac{5}{3}$
So, normal intersects x-axis at $\left(\frac{5}{3},0\right)$

So, area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}3& 2& 1\\ 6& 0& 1\\ \frac{5}{3}& 0& 1\end{array}\right|$
$=\frac{13}{3}$ sq.units