VITEEE 2013 Paper

Let P($x_1, y_1$) be a point on the ellipse $\frac{x^2}{18} + \frac{y^2}{32}$ = 1 ⇒ $\frac{x_1^2}{18} + \frac{y_1^2}{32}$ = 1 ... (i) The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{18} + \frac{yy_1}{32}$ = 1. This meets the axes at A $\left(\frac{18}{x_1}, 0\right)$ and B $\left(0, \frac{32}{y_1}\right)$. It is given that slope of the tangent at $(x_1, y_1)$ is - $\frac{4}{3}$ So , $-\frac{x_1}{18} \cdot \frac{32}{y_1}$ = - $\frac{4}{3}$ ⇒ $\frac{x_1}{y_1}$ = $\frac{3}{4}$ ⇒ $\frac{x_1}{3}$ = $\frac{y_1}{4}$ = K (say) ∴ $x_1$ = 3K and $y_1$ = 4K Putting $x_1, y_1$ in (i), we get $K^2$ = 1 ∴ Area of Δ OAB = $\frac{1}{2}$ OA . OB = $\frac{1}{2} \cdot \frac{18}{x_1} \cdot \frac{32}{y_1}$ = $\frac{1}{2} \frac{(18)(32)}{(3K)(4K)}$ = $\frac{24}{K^2}$ = 24 sq units (Since $K_2$ =1)