Let a line L_1 be tangent to the hyperbola x^216-y^24=1, and let L_2 be the line through t…

Question

Let a line $L_1$ be tangent to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}4=1$, and let $L_2$ be the line through the origin perpendicular to $L_1$. If the locus of the point of intersection of $L_1$ and $L_2$ is $(x^2+y^2)^2=\alpha x^2+\beta y^2$, then $\alpha+\beta$ is equal to\_\_\_\_

Answer: 12

Accepted Answer

For a tangent $lx+my=1$ to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$: $a^2l^2-b^2m^2=1$. Since $L_2$ (through the origin, perpendicular to $L_1$) has direction along $(l,m)$, the intersection point $P=(x,y)$ satisfies $l=\dfrac{x}{x^2+y^2},\ m=\dfrac{y}{x^2+y^2}$.

Substituting: $a^2x^2-b^2y^2=(x^2+y^2)^2$, i.e. $(x^2+y^2)^2=16x^2-4y^2$.

$$\alpha=16,\ \beta=-4,\qquad \alpha+\beta=12.$$

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