Find the radius of the largest circle which is passing through the focus of parabola y^2=4x and inscribed in the same parabola??

Ajith A M
15 Points
13 years ago

Ans  ::    2+sqrt(5)

centre of circle at (h,0).

so eqn of circle  (x-h)^2+y^2=r^2

substitute for h

solve the circle & parabola .....apply the condition for roots to be equal for the quadratic eqn thus obtained.

& then obtain r=2+5^0.5

yash
10 Points
9 years ago
Let (h,0) be the circle's center.
h=1+r
Equation of circle is:
(x-h)2+y2=r2
(x-1-r)2+y2=r2
x2+y2-2x-2xr+2r+1=0
But y2=4x, Substituting we get:
x2+2x-2xr+2r+1=0
x2+x(2-2r)+2r+1=0
Ax2+Bx+C=0 has equal roots since the circle intersects the parabola....
B2-4AC=0
(2-2r)2=4(2r+1)
r=0 or 4
r cannot be zero. Hence r=4.
yash
10 Points
9 years ago
Let (h,0) be the circle's center.
h=1+r
Equation of circle is:
(x-h)2+y2=r2
(x-1-r)2+y2=r2
x2+y2-2x-2xr+2r+1=0
But y2=4x, Substituting we get:
x2+2x-2xr+2r+1=0
x2+x(2-2r)+2r+1=0
Ax2+Bx+C=0 has equal roots since the circle intersects the parabola....
B2-4AC=0
(2-2r)2=4(2r+1)
r=0 or 4
r cannot be zero. Hence r=4.
yash
10 Points
9 years ago
Let (h,0) be the circle's center.
h=1+r
Equation of circle is:
(x-h)2+y2=r2
(x-1-r)2+y2=r2
x2+y2-2x-2xr+2r+1=0
But y2=4x, Substituting we get:
x2+2x-2xr+2r+1=0
x2+x(2-2r)+2r+1=0
Ax2+Bx+C=0 has equal roots since the circle intersects the parabola....
B2-4AC=0
(2-2r)2=4(2r+1)
r=0 or 4
r cannot be zero. Hence r=4.