Question

The tangent and normal to the ellipse $3x^2+5y^2=32$ at the point $P(2,2)$ meet the $x-$axis at $Q$ and $R$, respectively. Then the area (in sq. units) of the triangle $PQR$ is :

• A. $\frac{34}{15}$
• B. $\frac{16}{3}$
• C. $\frac{14}{3}$
• D. $\frac{68}{15}$

Answer 1

The correct option is D $\frac{68}{15}$

Given,
Equation of ellipse $\Rightarrow 3x^2+5y^2=32$
Diffrentiating both sides
$6x+10y{y}^{\prime }=0$
$y^{\prime }=\frac{-3x}{5y}$
$y_{(2,2)}^{\prime }=-\frac{3}{5}$

Equation of tangent at $(2,2)$ is,
$(y-2)=-\frac{3}{5}(x-2)$
$\Rightarrow 3x+5y=16$
$Q=(\frac{16}{3},0)$

Equation of normal at $(2,2)$ is,
$(y-2)=\frac{5}{3}(x-2)$
$\Rightarrow 5x-3y=4$
$R=(\frac{4}{5},0)$

$QR=\sqrt{{(\frac{16}{3}-\frac{4}{5})}^2}=\sqrt{{\left(\frac{68}{15}\right)}^2}=\frac{68}{15}$ and height $=2$

$\therefore \text{Area}=\frac{1}{2}\times \left(QR\right)\times 2=QR=\frac{68}{15}$