A circle of radius 2 units passes through the vertex and the focus of the parabola y^2=2x …

Question

A circle of radius $2$ units passes through the vertex and the focus of the parabola $y^2=2x$ and touches the parabola $y=\left(x-\dfrac14\right)^2+\alpha$, where $\alpha>0$. Then $(4\alpha-8)^2$ is equal to\_\_\_

Answer: 63

Accepted Answer

Vertex $(0,0)$, focus $\left(\tfrac12,0\right)$. Circle through both: $x\left(x-\tfrac12\right)+y^2+\lambda y=0$.

Radius: $\sqrt{\tfrac1{16}+\tfrac{\lambda^2}4}=2\Rightarrow\lambda^2=\dfrac{63}4$, centre $\left(\tfrac14,-\tfrac\lambda2\right)$.

Tangency with the parabola (touching along the common axis $x=\tfrac14$) gives $\alpha=-\dfrac\lambda2+2\Rightarrow(\alpha-2)^2=\dfrac{\lambda^2}4=\dfrac{63}{16}$.

$$(4\alpha-8)^2=16(\alpha-2)^2=63.$$

Solutions · 1