If the equation of the hyperbola with foci (4,2) and (8,2) is 3x^2-y^2- x+ y+=0, then ++ i…

Question

If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3x^2-y^2-\alpha x+\beta y+\gamma=0$, then $\alpha+\beta+\gamma$ is equal to \_\_\_\_\_\_\_\_.

Answer: 141

Accepted Answer

Centre $(6,2)$; write $\dfrac{(x-6)^2}{a^2}-\dfrac{(y-2)^2}{b^2}=1$, expanding to $b^2x^2-a^2y^2-12b^2x+4a^2y+36b^2-4a^2-a^2b^2=0$. Comparing with $3x^2-y^2-\ldots$: $\dfrac{b^2}{a^2}=3$.

$$e^2=1+\frac{b^2}{a^2}=4 \Rightarrow e=2.$$

Foci distance $2c=4=2ae \Rightarrow a=1,\ b^2=3$.

$$(x-6)^2-\frac{(y-2)^2}{3}=1 \;\Rightarrow\; 3x^2-y^2-36x+4y+101=0.$$

So $\alpha=36,\ \beta=4,\ \gamma=101$, and $\alpha+\beta+\gamma=141$.

Solutions · 1