Question:

<p>Equation of given ellipse is&nbsp;3x² + 5y² = 32&nbsp;...(i)<br /> Now, the slope of tangent and normal at point P(2, 2) to the ellipse (i) are respectively<br /> <span class="math-tex">$m_{T}=\frac{d y}{d x}|_{2,2} \text { and } m_{N}=-\left.\frac{d x}{d y}\right|_{2,2}$</span><br /> On differentiating ellipse (i), w.r.t. x, we get<br /> <span class="math-tex">$6 x+10 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{3 x}{5 y}$</span><br /> So,&nbsp;<span class="math-tex">$m_{T}=-\left.\frac{3x}{5 y}\right|_{(2,2)}=-\frac{3}{5} \text { and } m_{N}=\left.\frac{5 y}{3 y}\right|_{(2,2)}=\frac{5}{3}$</span><br /> Now, equation of tangent and normal to the given<br /> ellipse (i) at point P(2, 2) are<br /> <span class="math-tex">$(y-2)=-\frac{3}{5}(x-2)$</span><br /> and&nbsp;<span class="math-tex">$(y-2)=\frac{5}{3}(x-2)$</span>&nbsp;respectively.<br /> It is given that point of intersection of tangent and normal are Q and R at X-axis respectively.<br /> So,&nbsp;<span class="math-tex">$Q\left(\frac{1}{36}, 0\right) \text { and } R\left(\frac{4}{5}, 0\right)$</span><br /> <span class="math-tex">$\therefore \text { Area of } \Delta P Q R=\frac{1}{2}(Q R) \times \text { height }$</span><br /> <span class="math-tex">$=\frac{1}{2} \times \frac{68}{15} \times 2=\frac{68}{15} \text { sq units }$</span><br /> [<span class="math-tex">$\because$</span>&nbsp;<span class="math-tex">$Q R=\sqrt{\left(\frac{16}{3}-\frac{4}{5}\right)^{2}}=\sqrt{\left(\frac{68}{15}\right)^{2}}=\frac{68}{15}$</span> andheight = 2]</p>