A tangent having slope& - (4)/(3) to the ellipse x^ 2 18 + y^ 2 32...

Any point on the ellipse $\dfrac{x^{2}}{{ \lleft(3\sqrt{2})}^{2}} + \dfrac{y^{2}}{{ \left(4\sqrt{2}\right)}^{2}} =1$ can be taken as \left(3\sqrt{2}cos \theta,$4\sqrt{2}$sin \theta\right)$and the slope of the tangent as~$- $\dfrac{b^{2}x}{a^{2}y}$$<br /><br />Hence, slope$= - \dfrac{32 \lleft(3\sqrt{2}cos \theta)}{18 \l$left(4\sqrt{2}$sin $\theta)$} = $-\dfrac{4}{3}$cot \theta \ldots . \left(1\right)$<br /><br />Given, the slope of the tangent~$= - $$\dfrac{4}{3} \ldots .$$\left(2\right)

From equations (1) and (2), we get,~

$cot$\theta =$1 ⇒$\theta = \dfrac{\pi}{4}

Hence, the equation of the tangent is \dfrac{x . \dfrac{1}{\sqrt{2}}}{3\sqrt{2}} + \dfrac{y . \dfrac{1}{\sqrt{2}}}{4\sqrt{2}} =$1$

(i.e.) \dfrac{x}{6} + \dfrac{y}{8} =$1$~

Hence,& A = \left(6, 0\right), B = \left(0, 8\right)$<br /><br />Area of$\Delta OAB = \dfrac{1}{2} \times6 \times8 = 24sq . units