Single Choice

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

A$$a^2 + b^2$$
B$$\dfrac{(a + b)^2}{2}$$
C$$ab$$
Correct Answer
D$$\dfrac{(a - b)^2}{2}$$

Solution

Equation of tangent at $$(a cos \theta, b sin \theta)$$ to the ellipse is
$$\dfrac{x}{a} cos \theta + \dfrac{y}{b} sin \theta = 1$$
Coordinates of P and Q are
$$\left( \dfrac{a}{cos \theta} , 0 \right )$$ and $$\left( 0 , \dfrac{b}{sin \theta}\right )$$, respectively.
Area of $$\Delta OPQ = \dfrac{1}{2} \left| \dfrac{a}{cos \theta} \times \dfrac{b}{sin \theta} \right | = \dfrac{ab}{|sin 2 \theta|}\geq ab$$ .......... [$$\because |sin \theta|\leq 1$$]
$$\therefore$$ Minimum area $$= ab$$
Hence, option C is correct.

SIMILAR QUESTIONS

Circles

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

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 $$\displaystyle \frac{8}{\sqrt{5}}\Delta_{1}-8\Delta_{2}=$$

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