A tangent having slope of -frac{4}{3} to the | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$. Since $b^2 > a^2$,the major axis is along th

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$24$

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(B) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$. Here $a^2 = 18$ and $b^2 = 32$. Since $b^2 > a^2$,the major axis is along the $y$-axis.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 slope $m = -\frac{4}{3}$,the equation of the tangent is $y = -\frac{4}{3}x \pm \sqrt{18(-\frac{4}{3})^2 + 32} = -\frac{4}{3}x \pm \sqrt{18(\frac{16}{9}) + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm 8$.Taking the positive intercept case,$y = -\frac{4}{3}x + 8$,which simplifies to $4x + 3y = 24$,or $\frac{x}{6} + \frac{y}{8} = 1$.The tangent intersects the $x$-axis at $A(6, 0)$ and the $y$-axis at $B(0, 8)$. The centre $C$ is $(0, 0)$.The area of triangle $ABC$ is $\frac{1}{2} 	imes |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = \frac{1}{2} 	imes |6(8 - 0) + 0(0 - 0) + 0(0 - 8)| = \frac{1}{2} 	imes 48 = 24$ $sq. \,units$.

$A$ 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. If $C$ is the centre of the ellipse,then the area of the triangle $ABC$ is: .............. $sq. \,units$

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