Let A be a point on the x-axis.

Question

Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(QR)^2$ is equal to:

Accepted Answer

Tangent to $y^2=16x$ ($a=4$): $y=mx+\dfrac4m$. Tangent to the circle $x^2+y^2=8$ (radius $2\sqrt2$) requires
$$\frac{|4/m|}{\sqrt{1+m^2}}=2\sqrt2\ \Longrightarrow\ m^4+m^2-2=0\ \Longrightarrow\ m^2=1\ \Longrightarrow\ m=\pm1.$$

Take $m=1$: line $x-y+4=0$. Point of contact on the parabola: $R=(a/m^2,2a/m)=(4,8)$. Point of contact on the circle (foot of perpendicular from origin): $Q=(-2,2)$.

$(QR)^2=(4+2)^2+(8-2)^2=36+36=72.$

Solutions · 1