Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.
My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation for pair of tangents is $SS_1=T^2$, where $S_1=0$ is obtained after putting the external point in $S=0$. And $T=0$ is obtained after changing $x^2$ to $xx_1$ and $y^2$ to $yy_1$ in $S=0$, where $(x_1,y_1)$ is the external point from which the tangents are being drawn.
So,
$$SS_1=T^2$$
$$\implies (x^2+3y^2-9)(\alpha^2+3\alpha^4-9)=(x\alpha+3y\alpha^2-9)^2$$
Now, if $\theta$ is the angle between the tangents then $\tan\theta=|\frac{2\sqrt{h^2-ab}}{a+b}|$, where, $2h=$ coefficient of $xy=6\alpha^3, a=$ coefficient of $x^2=3\alpha^4-9$ and $b=$ coefficient of $y^2=3\alpha^2-27$
Therefore, $\tan\theta=|\frac{2\sqrt{9\alpha^6-9(\alpha^4-3)(\alpha^2-9)}}{3\alpha^4+3\alpha^2-36}|=|\frac{2\sqrt{\alpha^6-(\alpha^6-9\alpha^4-3\alpha^2+27)}}{\alpha^4+\alpha^2-12}|=|\frac{2\sqrt{9\alpha^4+3\alpha^2-27}}{\alpha^4+\alpha^2-12}|$
Not able to proceed next.
Best Answer
Hint: $\alpha^4+\alpha^2-12=(\alpha^2-3)(\alpha^2+4)$ making the angle $\pi/2$ for $\alpha=\pm\sqrt{3}.$