What is the largest value of a for which the circle x^2+y^2=a^2 lies completely inside the parabola y^2=4(x+4) ?

1 Answer
Jan 28, 2018

a=2\sqrt{3}.

Explanation:

Let's work this out. We have

y^2=a^2-x^2

y^2=4(x+4)

So then when the circle intersects the parabola the values of y^2 must agree:

a^2-x^2=4(x+4)

x^2+4x+(16-a^2)=0

We now find the positive value of a where the quadratic equation has just one root, where the circle will just touch the parabola. To get that the discriminant must be zero:

(Middle coefficient)^2-(4×(product of other coefficients))=0

4^2-(4×(16-a^2))=0

a^2=12=2^2×3

a=2\sqrt{3}.