The tangent to the ellipse 9x^2 + 16y^2 = 288 | vedclass

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(A) The equation of the ellipse is $9x^2 + 16y^2 = 288$. Dividing by $288$, we get $\frac{x^2}{32} + \frac{y^2}{18} = 1$. Here $a^2 = 32$ and $b^2 = 1

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$25$ sq. units

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(A) The equation of the ellipse is $9x^2 + 16y^2 = 288$. Dividing by $288$, we get $\frac{x^2}{32} + \frac{y^2}{18} = 1$. Here $a^2 = 32$ and $b^2 = 18$.The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Since the tangent makes equal intercepts on the axes, its slope must be $m = -1$ or $m = 1$. Given it makes intercepts on the axes, we consider the form $\frac{x}{c} + \frac{y}{c} = 1$, which implies $y = -x + c$, so $m = -1$.The condition for tangency is $c^2 = a^2m^2 + b^2$. Substituting $m = -1$, $c^2 = 32(-1)^2 + 18 = 32 + 18 = 50$.Thus, $c = \pm \sqrt{50} = \pm 5\sqrt{2}$.The intercepts are $A(5\sqrt{2}, 0)$ and $B(0, 5\sqrt{2})$.The area of $	riangle OAB = \frac{1}{2} 	imes |5\sqrt{2}| 	imes |5\sqrt{2}| = \frac{1}{2} 	imes 50 = 25$ sq. units.

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the co-ordinate axes intersects the $X$-axis and the $Y$-axis in the points $A$ and $B$ respectively. Then $\text{Area}(\triangle OAB) = $ (where $O$ is origin)

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