CBSE BoardEnglish MediumSTD 11 ScienceMathsConic Sections3 Marks
Question
Find the equation of the hyperbola with:
$(0,\pm\sqrt{10}) $passing through $(2,3)$
✓
Answer
Given that, $\text{foci}=(0,\pm\sqrt{10})$
$\therefore\text{ be}=\sqrt{10}$
Also, $\text{a}^2=\text{b}^2(\text{e}^2-1)$
$\Rightarrow\text{a}^2=\text{b}^2\text{e}^2-\text{b}^2=10-\text{b}^2$
$\therefore$ Equation of the hyperbola is,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=-1$ or $\frac{\text{x}^2}{10-\text{b}^2}-\frac{\text{y}^2}{\text{b}^2}=-1$
Since, hyperbola passes through the point $(2, 3)$
$\therefore\ \frac{4}{10-\text{b}^2}-\frac{9}{\text{b}^2}=-1$
$\Rightarrow4\text{b}^2-9(10-\text{b}^2)=-\text{b}^2(10-\text{b}^2)$
$\Rightarrow\text{b}^4-23\text{b}^2+90=0$
$\Rightarrow(\text{b}^2-18)(\text{b}^2-5)=0$
$\Rightarrow\text{b}^2=5 (b^2 = 18$ not possible as $a^2 + b^2 = 10)$
$\therefore\ \text{b}^2=10-5=5$
So, the equation of hyperbola is $\frac{\text{x}^2}{5}-\frac{\text{y}^2}{5}=-1$ or $y^2 - x^2 = 5.$
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