Let a and b be any two numbers satisfying 1 a 2 + 1 b 2 = 1 4 . Then, the foot of perpendicular from the…

Let a and b be any two numbers satisfying 1a2+1b2=14. Then, the foot of perpendicular from the origin on the variable line xa+yb=1 lies on :
  1. A circle of radius =2
  2. A hyperbola with each semi-axis = 2 .
  3. A hyperbola with each semi-axis =2
  4. A circle of radius = 2

Solution

The given line is bx+ay=ab.

Foot of perpendicular from origin

xb=ya=--aba2+b2

x=ab2a2+b2, y=a2ba2+b2

x2+y2=a2b2a2+b2a2+b22

x2+y2=a2b2a2+b2

given 1a2+1b2=14

 a2+b2a2b2=14

⇒  x2+y2=4, which is equation of circle.
 

Asked in: JEE Main 2014 (09 Apr Online)

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