If a tangent with slope -4/3 to the ellipse f | 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 with slope $m$ to t

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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 with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given $m = -4/3$,$a^2 = 18$,and $b^2 = 32$.Substituting these values: $y = -\frac{4}{3}x \pm \sqrt{18(-\frac{4}{3})^2 + 32} = -\frac{4}{3}x \pm \sqrt{18(\frac{16}{9}) + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm \sqrt{64} = -\frac{4}{3}x \pm 8$.Taking the positive intercept case,$y = -\frac{4}{3}x + 8$.To find $A$ (intersection with the major axis,which is the $y$-axis here since $b^2 > a^2$),set $x = 0$: $y = 8$,so $A = (0, 8)$.To find $B$ (intersection with the minor axis,which is the $x$-axis),set $y = 0$: $0 = -\frac{4}{3}x + 8 \implies \frac{4}{3}x = 8 \implies x = 6$,so $B = (6, 0)$.The area of $\Delta OAB = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes 8 	imes 6 = 24$ square units.

If a tangent with slope $-4/3$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major axis and minor axis at $A$ and $B$ respectively,then the area of $\Delta OAB$ is .......... square units.

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