Question

Referred to the principal axes as the axes of coordinates find the equation of the hyperbola whose foci are at $(0,\pm \sqrt{10})$ and which passes through the point $(2,3)$.

• A. $x^2-y^2=1$
• B. $x^2-y^2=-5$
• C. None of the above
• D. $x^2+y^2=1$

Answer 1

The correct option is B $x^2-y^2=-5$

Since the vertices are on $y$-axis, so let the equation of the required hyperbola be

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ $\cdots \left(i\right)$

It passes through $(2,3)$.

$\therefore \frac{4}{a^2}-\frac{9}{b^2}=-1\Rightarrow \frac{4}{b^2(e^2-1)}-\frac{9}{b^2}=-1$
$\Rightarrow \frac{4}{b^2{e}^2-b^2}-\frac{9}{b^2}=-1$ $\cdots \left(ii\right)$

$[\because a^2=b^2(e^2-1)]$

The coordinates of foci are $(0,\pm \sqrt{10})$

$\therefore be=\sqrt{10}\Rightarrow b^2{e}^2=10$ $\cdots \left(iii\right)$

From (ii) and (iii), we get

$\frac{4}{10-b^2}-\frac{9}{b^2}=-1\Rightarrow 4b^2-9(10-b^2)=-b^2(10-b^2)$

$\Rightarrow 13b^2-90=-10b^2+b^4\Rightarrow b^4-23b^2+90=0$

$\Rightarrow (b^2-18)(b^2-5)=0\Rightarrow b^2=18$ or $b^2=5$.

If $b^2=18$, then $a^2=10-18=-8$, which is not possible.

$b^2=5\Rightarrow a^2=10-5=5.$

Substituting the values of $a^2$ and $b^2$ in (i), we get $\frac{x^2}{5}-\frac{y^2}{5}=-1$ ie. $x^2-y^2=-5$ as the equation of the required hyperbola.