Gujarat BoardEnglish MediumSTD 11 ScienceMATHSHyperbola4 Marks
Question
Find the equation of the hyperbola whose Focus is at $(4, 2)$ centre at $(6, 2)$ and $e = 2$.
✓
Answer
The equation of the hyperbola with centre ($X_0, Y_0$_) is given by $\frac{\text{(x}-\text{x}_0)}{\text{a}^{2}}-\frac{(\text{y}-\text{y}_0)^{2}}{\text{b}^{2}}=1$ Focus = $(\text{ae}+\text{x}_0, \text{y}_0)$
$\therefore\text{ae} = -2$
$\Rightarrow\text{a} = -1$
$\text{b}^{2} = (-2)^{2}-\text{a}^{2}$
$\Rightarrow\text{b}^{2}=(-2)^{2}-(-1)^{2}$
$\Rightarrow\text{b}^{2}=3$
$\Rightarrow\frac{\text{(x}-6)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$
$\Rightarrow3\text{(x}-6)-(\text{y}-2)^{2}=3$
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