The area (in sq. units) of the quadrilateral | vedclass

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(D) Given ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$. Here $a^2=16$ and $b^2=12$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{12}{16}

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(D) Given ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$. Here $a^2=16$ and $b^2=12$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$. The end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 4 	imes \frac{1}{2}, \pm \frac{12}{4}) = (\pm 2, \pm 3)$. The equation of the tangent at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$. For the point $(2, 3)$,the tangent is $\frac{2x}{16}+\frac{3y}{12}=1$,which simplifies to $\frac{x}{8}+\frac{y}{4}=1$. This line intersects the $x$-axis at $(8, 0)$ and the $y$-axis at $(0, 4)$. Due to symmetry,the quadrilateral formed by the four tangents at $(\pm 2, \pm 3)$ is a rhombus with vertices at $(\pm 8, 0)$ and $(0, \pm 4)$. The area of this rhombus is $4 	imes (	ext{Area of triangle with vertices } (0,0), (8,0), (0,4)) = 4 	imes (\frac{1}{2} 	imes 8 	imes 4) = 64$ sq. units.

The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S \equiv \frac{x^2}{16}+\frac{y^2}{12}=1$ is

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