Question

Let the equation of the hyperbola with foci \((1, 5), (1, -1)\) and eccentricity \(\sqrt{3}\) be \(x^2 - 2y^2 + ax + by + c = 0\). Then \(|a + b + c|\) is equal to ________.

Hint:

The foci lie on the vertical line \(x=1\), so the transverse axis is vertical with centre \((1,2)\); find \(a^2\) and \(b^2\) from \(ae=3\).

Answer 1

Correct Answer: 5

Use the focus-directrix property. Centre \((1,2)\), \(c=3\), \(a=c/e=\sqrt3\), so the directrices are \(y=2\pm\dfrac{a}{e}=2\pm1\); the directrix for the focus \((1,5)\) is \(y=3\).
For a point \((x,y)\) on the hyperbola: distance to focus \(=e\times\) distance to directrix, so \((x-1)^2+(y-5)^2=3(y-3)^2\).
Expand: \(x^2-2x+1+y^2-10y+25=3y^2-18y+27\Rightarrow x^2-2y^2-2x+8y-1=0\).
Hence \(a=-2,\ b=8,\ c=-1\) and \(|a+b+c|=5\).
Answer: 5