P (x, y) is a point on a straight line which makes intercepts a and b on the x, y axes respectively, then the minimum value of $x^{2} + y^{2} =$

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

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

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

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

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Solution

The correct option is C $\frac{a^{2} b^{2}}{(a^{2} + b^{2})}$
$s l o p e = \frac{- b}{a} = \frac{y - b}{x}$
$- b x = a y - a b$
$f (x) = x^{2} + (\frac{a b - b x}{a})^{2}$
$f^{'} (x) = 2 x + \frac{2 b^{2}}{a^{2}} (a - x) (- 1)$
$0 = x - \frac{b^{2}}{a^{2}} (a - x)$
$\frac{b^{2}}{a} = \frac{(a^{2} + b^{2})}{a^{2}} x$
$x = \frac{a b^{2}}{(a^{2} + b^{2})}$
$y = \frac{b a^{2}}{(a^{2} + b^{2})}$
$(x^{2} + y^{2})_{m i n} = \frac{a^{2} b^{2}}{(a^{2} + b^{2})}$

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The length of the chord intercepted by the circle $x^{2} + y^{2} = r^{2}$ on the line $\frac{x}{a} + \frac{y}{b} = 1$

The equation of the circle with center (– a, – b) and radius √a² – b² is:

x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}}

a^{2} - b^{2} = \frac{2 a^{2}}{x^{2} + 1}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

p

q

\frac{a^{2}}{p^{2}} + \frac{b^{2}}{q^{2}} =

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