Question

The equation of transverse and conjugate axis of hyperbola are $x+2y-3=0,2x-y+4=0$ respectively, and their respective lengths are $\sqrt{2}$ and $\frac{2}{\sqrt{3}}$. The equation of hyberbola is

• A. $\frac{2}{5}(x+2y-3{)}^2-\frac{3}{5}(2x-y+4{)}^2=1$
• B. $\frac{2}{5}(2x-y+4{)}^2-\frac{3}{5}(x+2y-3{)}^2=1$
• C. $2(2x-y+4{)}^2-3{(x+2y-3)}^2=1$
• D. $2(x+2y-3{)}^2-3(2x-y+4{)}^2=1$

Answer 1

The correct option is B $\frac{2}{5}(2x-y+4{)}^2-\frac{3}{5}(x+2y-3{)}^2=1$

Equation of hyperbola described on two perpendicular lines is :
$\frac{(PN{)}^2}{a^2}-\frac{\left(P{M}^2\right)}{b^2}=1$
Where $PN$ is perpendicular distance from point on hyperbola to conjugate axis.
And $PM$ is perpendicular distance from point on hyperbola to transverse axis.
$PN=\left|\frac{2x-y+4}{\sqrt{5}}\right|$ and $PM=\left|\frac{x+2y-3}{\sqrt{5}}\right|$
Required equation of hyperbola is :
$\frac{{\left(\frac{2x-y+4}{\sqrt{5}}\right)}^2}{{\left(\frac{1}{\sqrt{2}}\right)}^2}-\frac{{\left(\frac{x+2y-3}{\sqrt{5}}\right)}^2}{{\left(\frac{1}{\sqrt{3}}\right)}^2}=1$
$\therefore \frac{2}{5}(2x-y+4{)}^2-\frac{3}{5}(x+2y-3{)}^2=1$