If two tangents drawn from a point (,) lying on the ellipse 25x^2+4y^2=1 to the parabola y…

Question

If two tangents drawn from a point $(\alpha,\beta)$ lying on the ellipse $25x^2+4y^2=1$ to the parabola $y^2=4x$ are such that the slope of one tangent is four times the other, then the value of $(10\alpha+5)^2+(16\beta^2+50)^2$ equals:

Answer: 2929

Accepted Answer

Tangent to $y^2=4x$: $y=mx+1/m$; through $(\alpha,\beta)$: $m^2\alpha-\beta m+1=0$. With roots $m,4m$:
$$5m=\frac\beta\alpha,\qquad 4m^2=\frac1\alpha\ \Longrightarrow\ 4\beta^2=25\alpha.$$

On the ellipse: $25\alpha^2+4\beta^2=1\Rightarrow25\alpha^2+25\alpha-1=0\Rightarrow\alpha=\dfrac{-5+\sqrt{29}}{10}$ (taking the root giving a valid point).

Then $10\alpha+5=\sqrt{29}$, so $(10\alpha+5)^2=29$.

Also $16\beta^2=4(4\beta^2)=4(25\alpha)=100\alpha$, so $16\beta^2+50=100\alpha+50=10\sqrt{29}$, giving $(16\beta^2+50)^2=2900$.

$$(10\alpha+5)^2+(16\beta^2+50)^2=29+2900=2929.$$

Solutions · 1