Two common tangents to the circle x^2+y^2=2a^2 and the parabola y^2=8ax are:

Question

Two common tangents to the circle $x^2+y^2=2a^2$ and the parabola $y^2=8ax$ are:

Accepted Answer

Tangent to $y^2=8ax$: $y=mx+\dfrac{2a}{m}$. For tangency to $x^2+y^2=2a^2$ (radius $a\sqrt2$), the perpendicular distance from the origin equals the radius:
$$\frac{2a/m}{\sqrt{1+m^2}}=a\sqrt2\ \Rightarrow\ \frac{4}{m^2}=2(1+m^2)\ \Rightarrow\ m^4+m^2-2=0\ \Rightarrow\ (m^2+2)(m^2-1)=0\ \Rightarrow\ m=\pm1.$$

Substituting back: $y=\pm x\pm2a$, i.e. $y=\pm(x+2a).$

Solutions · 1