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 :

Correct answer: A. A circle of radius = 2 The given line is bx + ay = ab . Foot of perpendicular from origin (x)/(b) = (y)/(a) = - [ - ab 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.

JEE Main 20149 AprilOnline

Exam: JEE Main
Year: 2014
Date: 9 April
Mode: Online
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2014 · 9 April · Online · Mathematics · Straight Lines

Let a and b be any two numbers satisfying 1 a ^ 2 + 1 b ^ 2 =...

Let

a

and

b

be any two numbers satisfying

\dfrac{1}{{a}^{2}} + \dfrac{1}{{b}^{2}} = \dfrac{1}{4}.

Then, the foot of perpendicular from the origin on the variable line

\dfrac{x}{a} + \dfrac{y}{b} = 1

lies on :

Options

AA circle of radius $= 2$
BA hyperbola with each semi-axis $= \sqrt{2}$ .
CA hyperbola with each semi-axis $= 2$
DA circle of radius $= \sqrt{2}$

Correct answer

A. A circle of radius $= 2$

Step-by-step solution

The given line is

bx + ay = ab

.

Foot of perpendicular from origin

\dfrac{x}{b} = \dfrac{y}{a} = - \left[\dfrac{- ab}{{a}^{2} + {b}^{2}}\right]

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

{x}^{2} + {y}^{2} = \dfrac{{a}^{2}{b}^{2}\left({a}^{2} + {b}^{2}\right)}{{\left({a}^{2} + {b}^{2}\right)}^{2}}

{x}^{2} + {y}^{2} = \dfrac{{a}^{2}{b}^{2}}{{a}^{2} + {b}^{2}}

given

\dfrac{1}{{a}^{2}} + \dfrac{1}{{b}^{2}} = \dfrac{1}{4}

\dfrac{{a}^{2} + {b}^{2}}{{a}^{2}{b}^{2}} = \dfrac{1}{4}

⇒ {x}^{2} + {y}^{2} = 4

, which is equation of circle.

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