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 = 18

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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 $m$ must be $\pm 1$. Given it makes equal intercepts,the line equation is $x + y = c$ or $x - y = c$. For equal intercepts on the axes,the slope is $-1$,so the tangent is $y = -x + c$,or $x + y = c$.The condition for $y = mx + c$ to be a tangent is $c^2 = a^2m^2 + b^2$. Substituting $m = -1$,$a^2 = 32$,and $b^2 = 18$:$c^2 = 32(-1)^2 + 18 = 32 + 18 = 50$.Thus,$c = \pm 5\sqrt{2}$.The intercepts are $A(c, 0)$ and $B(0, c)$. The area of $	riangle OAB$ is $\frac{1}{2} |c| |c| = \frac{1}{2} c^2 = \frac{1}{2} (50) = 25$ sq. units.

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the coordinate axes intersects the $X$-axis and the $Y$-axis at points $A$ and $B$ respectively. Then,the area of $\triangle OAB$ (where $O$ is the origin) is:

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