The area of the quadrilateral formed by the t | vedclass

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(d) By symmetry the quadrilateral is a rhombus.

So area is four times the area of the right angled triangle formed by the tangent and axes in the I

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(d) By symmetry the quadrilateral is a rhombus.

So area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant.

Now, $ae = \sqrt {{a^2} - {b^2}} $

$\Rightarrow ae = 2$

==> Tangent (in first quadrant) at end of latus rectum $\left( {2,\frac{5}{3}} \right)$ is

$\frac{2}{9}x + \frac{5}{3}\frac{y}{5} = 1$

$i.e.$, $\frac{x}{{9/2}} + \frac{y}{3} = 1$

Area $ = 4.\,\frac{1}{2}.\,\frac{9}{2}.3 = 27$ $sq.\ unit.$

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$, is .............. $\mathrm{sq. \,units}$

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