Single Choice

The area (in sq. units) of the quadrilateral formed by the tangent at the end points of the latera recta to the ellipse $$\displaystyle \dfrac {x^2}{9}+\dfrac {y^2}{5}=1$$, is

A$$\dfrac {27}{4}$$
B$$18$$
C$$\dfrac {27}{2}$$
D$$27$$
Correct Answer

Solution

For the given ellipse

$$e = \sqrt { 1 - \dfrac{5}{9} } = \dfrac{2}{3} $$

The endpoint of one of the latus rectum will be $$ \left (2,\dfrac53 \right) $$

The tangents drawn at the end of the latera recta to the ellipse will intersect at the $$x$$ and the $$y$$ axis because of symmetry. The area of the quadrilateral thus formed will be $$4$$ times the area of the triangle formed by the tangent and the coordinate axis.

The equation of the tangent at $$\left(2,\dfrac{5}{3} \right) $$ can be written as :
$$ \dfrac{2x}{9} + \dfrac{5y}{15} = 1 $$

X intercept $$ =\dfrac{9}{2} $$

Y intercept $$ = 3 $$

Area of the triangle $$ = \dfrac12 \times \dfrac92 \times 3 =\dfrac{27}{4} $$

Hence, area of the quadrilateral $$ = 4 \times \dfrac{27}{4} = 27 $$

SIMILAR QUESTIONS

Circles

A vertical line passing through the point $$(\mathrm{h}, 0)$$ intersects the ellipse $$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. Let the tangents to the ellipse at $$\mathrm{P}$$ and $$\mathrm{Q}$$ meet at the point $$\mathrm{R}$$. If $$\Delta(\mathrm{h})=$$ area of the triangle $$\mathrm{P}\mathrm{Q}\mathrm{R},\ \Delta_{1}= \underset{1/2\leq h\leq 1}{max} \Delta(\mathrm{h})$$ and $$\displaystyle \Delta_{2}=\min_{1/2\leq h\leq 1}\Delta(\mathrm{h})$$ , then $$\displaystyle \frac{8}{\sqrt{5}}\Delta_{1}-8\Delta_{2}=$$

Circles

The minimum area of the triangle formed by any tangent to the ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with the coordinate axes is

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