If the tangent to the parabola y^2=x at a point (,), (0) is also a tangent to the ellipse…

Question

If the tangent to the parabola $y^2=x$ at a point $(\alpha,\beta)$, ($\beta>0$) is also a tangent to the ellipse $x^2+2y^2=1$, then $\alpha$ is equal to:

Accepted Answer

Tangent to $y^2=x$ with slope $m$: $y=mx+\dfrac1{4m}$. Tangent to $x^2+2y^2=1$: $y=mx\pm\sqrt{m^2+\dfrac12}$.

$$\frac1{4m}=\pm\sqrt{m^2+\frac12}\Rightarrow\frac1{16m^2}=m^2+\frac12\Rightarrow16m^4+8m^2-1=0\Rightarrow m^2=\frac{\sqrt2-1}4.$$

$$\alpha=\frac1{4m^2}=\frac1{4\cdot\frac{\sqrt2-1}4}=\frac1{\sqrt2-1}=\sqrt2+1.$$

Solutions · 1