Let a and b be any two numbers satisfying 1 a 2 + 1 b 2 = 1 4 . Then, the foot of perpendicular from the origin on the variable line x a + y b = 1 lies on :

The given line is b x + a y = a b . Foot of perpendicular from origin x b = y a = - - a b a 2 + b 2 x = a b 2 a 2 + b 2 , y = a 2 b a 2 + b 2 x 2 + y 2 = a 2 b 2 a 2 + b 2 a 2 + b 2 2 x 2 + y 2 = a 2 b 2 a 2 + b 2 given 1 a 2 + 1 b 2 = 1 4 a 2 + b 2 a 2 b 2 = 1 4 ⇒ x 2 + y 2 = 4 , which is equation of circle.

Mathematics - Point Question with Solution | TestHub

MathematicsPointLocusHard2 minPYQ_2014

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Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{4} .$ Then, the foot of perpendicular from the origin on the variable line $\frac{x}{a} + \frac{y}{b} = 1$ lies on :

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Answer:

Solution:

The given line is $b x + a y = a b$ .

Foot of perpendicular from origin

$\frac{x}{b} = \frac{y}{a} = - [\frac{- a b}{a^{2} + b^{2}}]$

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

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

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

given $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{4}$

$\frac{a^{2} + b^{2}}{a^{2} b^{2}} = \frac{1}{4}$

$\Rightarrow x^{2} + y^{2} = 4$ , which is equation of circle.

Stream:JEESubject:MathematicsTopic:PointSubtopic:Locus

⏱ 2mℹ️ Source: PYQ_2014

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