Let the circle $C$ touch the line $x-y+1=0$, have the centre on the positive $x$-axis, and cut off a chord of length $\frac4{\sqrt{13}}$ along the line $-3x+2y=1$. Let $H$ be the hyperbola $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$, whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2+3\beta^2$ is equal to ______

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Let the centre be $(h,0)$. Radius $r=\frac{h+1}{\sqrt2}$. Distance of the centre from $3x-2y+1=0$ is $\frac{3h+1}{\sqrt{13}}$, so $r^2=\frac{(3h+1)^2}{13}+\frac4{13}$. Then $13(h+1)^2=2(3h+1)^2+8\Rightarrow5h^2-14h-3=0$, $h=3$. $r^2=8$, so the diameter is $2r=4\sqrt2=2\alpha$ and $\alpha^2=8$. The focus $(3,0)$ gives $\alpha^2+\beta^2=9$, so $\beta^2=1$. $2\alpha^2+3\beta^2=16+3=19$. Answer: 19.

Explanation

Reviewed by Dr. Abubacker

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