JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q79

← JEE Main 2014 · 9 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 79 of 90

Let $a$ and $b$ be any two numbers satisfying $\dfrac1{a^2}+\dfrac1{b^2}=\dfrac14$ . Then, the foot of perpendicular from the origin on the variable line, $\dfrac xa+\dfrac yb=1$ , lies on :

a hyperbola with each semi-axis

=\sqrt2

a hyperbola with each semi-axis = 2.

a circle of radius = 2

a circle of radius

=\sqrt2

Answer and worked solution

Official answer: C

The distance from the origin to the line is
$p=\frac1{\sqrt{1/a^2+1/b^2}}=2.$
The foot $(h,k)$ of the perpendicular is at distance $p$ from the origin, so it lies on $h^2+k^2=4$ : a circle of radius 2.

Topics:Distance of a point from a line Locus

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 79