Minimum area of triangle by any tangent to ellipse frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}} =

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10-2. Parabola, Ellipse, Hyperbola

easy

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

$\frac{{{a^2} + {b^2}}}{2}$

$\frac{{{{(a + b)}^2}}}{2}$

$ab$

$\frac{{{{(a - b)}^2}}}{2}$

(IIT-2005)

Solution

$\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$

$P = \left( {\frac{a}{{\cos \theta }},\,0} \right)$

$Q = \left( {0,\,\frac{b}{{\sin \theta }}} \right)$

Area of $OPQ = \frac{1}{2}\left| {\,\left( {\frac{a}{{\cos \theta }}} \right)\,\left( {\frac{b}{{\sin \theta }}} \right)\,} \right| = \frac{{ab}}{{|\sin 2\theta |}}$

${({\rm{Area}})_{\min }} = ab$.

Standard 11

Mathematics

From the point$ C(0,\lambda )$ two tangents are drawn to ellipse $x^2\ +\ 2y^2\ = 4$ cutting major axis at $A$ and $B$. If area of $\Delta$ $ABC$ is minimum, then value of $\lambda$ is-

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(IIT-1968)

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hard

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normal

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

Solution

Similar Questions

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