If a tangent having slope of - frac{4}{3} to | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$.The equation of a tangent to the ellipse with

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$24$

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(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$.The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given slope $m = -\frac{4}{3}$,the equation of the tangent is $y = -\frac{4}{3}x \pm \sqrt{18\left(-\frac{4}{3}ight)^2 + 32}$.$y = -\frac{4}{3}x \pm \sqrt{18 	imes \frac{16}{9} + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm \sqrt{64} = -\frac{4}{3}x \pm 8$.Taking the positive case,$y = -\frac{4}{3}x + 8$,which can be written as $\frac{4}{3}x + y = 8$,or $\frac{x}{6} + \frac{y}{8} = 1$.The tangent intersects the axes at $A(6, 0)$ and $B(0, 8)$.The area of $\Delta OAB = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes 6 	imes 8 = 24 	ext{ sq. units}$.

If a tangent having slope of $- \frac{4}{3}$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major and minor axes in points $A$ and $B$ respectively,then the area of $\Delta OAB$ is equal to .................. $sq. \text{ units}$ ($O$ is the centre of the ellipse).

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