Tangent to the ellipse frac{x^{2}}{32}+frac{y | vedclass

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(C) The equation of the ellipse is $\frac{x^{2}}{32}+\frac{y^{2}}{18}=1$,where $a^{2}=32$ and $b^{2}=18$. The equation of a tangent with slope $m = -\

Vedclass · Accepted Answer

$24$ sq unit

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(C) The equation of the ellipse is $\frac{x^{2}}{32}+\frac{y^{2}}{18}=1$,where $a^{2}=32$ and $b^{2}=18$. The equation of a tangent with slope $m = -\frac{3}{4}$ is $y = mx \pm \sqrt{a^{2}m^{2}+b^{2}}$. Substituting the values: $y = -\frac{3}{4}x \pm \sqrt{32 	imes (-\frac{3}{4})^{2} + 18}$. $y = -\frac{3}{4}x \pm \sqrt{32 	imes \frac{9}{16} + 18} = -\frac{3}{4}x \pm \sqrt{18 + 18} = -\frac{3}{4}x \pm 6$. Considering the positive intercept,the equation is $y = -\frac{3}{4}x + 6$,which simplifies to $3x + 4y = 24$. The intercepts are found by setting $y=0$ and $x=0$: For $y=0$,$3x=24 \Rightarrow x=8$,so $A = (8, 0)$. For $x=0$,$4y=24 \Rightarrow y=6$,so $B = (0, 6)$. The area of $\Delta AOB = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes 8 	imes 6 = 24$ sq unit.

Tangent to the ellipse $\frac{x^{2}}{32}+\frac{y^{2}}{18}=1$ having slope $-\frac{3}{4}$ meets the coordinate axes at $A$ and $B$. Find the area of the $\Delta AOB$,where $O$ is the origin.

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