Question

The equations of the transverse and conjugate axes of a hyperbola are respectively $x+2y-3=0,2x-y+4=0$ and their respective lengths are $\sqrt{2}$ 2/$\sqrt{2}$. The equation of the hyperbola is

• A. $\frac{2}{5}(x+2y-3{)}^2-\frac{3}{5}(2x-y+4{)}^2=1$
• B. $\frac{2}{5}(2x+y-4{)}^2-\frac{3}{5}(x+2y3-4{)}^2=1$
• C. $2(2x-y+4{)}^2-3(x+2y-3{)}^2=1$
• D. $2(2x+2y-3{)}^2-3(2x-y+4{)}^2=1$

Answer 1

The correct option is B $\frac{2}{5}(2x+y-4{)}^2-\frac{3}{5}(x+2y3-4{)}^2=1$

It is given that $2a=\sqrt{2}$ which implies that $a=\frac{1}{\sqrt{2}}$.

Also, $2b=\frac{2}{\sqrt{3}}\Rightarrow b=\frac{1}{\sqrt{3}}$

If we take the two axes as the new coordinate system and the point of intersection of the axes of the new origin, then in the new coordinate system, equation of the hyperbola will be:

$\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1\Rightarrow \frac{X^2}{{\left(\frac{1}{\sqrt{2}}\right)}^2}-\frac{Y^2}{{\left(\frac{1}{\sqrt{3}}\right)}^2}=1\Rightarrow \frac{X^2}{\frac{1}{2}}-\frac{Y^2}{\frac{1}{3}}=1\Rightarrow 2X^2-3Y^2=1....\left(1\right)$

Let $P(x,y)$ be the coordinates of a point on the hyperbola in original x-y system, then

$X=\frac{|2x-y+4|}{\sqrt{5}},Y=\frac{|x+2y-3|}{\sqrt{5}}$

($\because$ $X$ is the distance of a point on hyperbola from $2x-y+4=0$ and $Y$ is the distance of a point on hyperbola from $x+2y-3=0$)

Therefore, equation 1 becomes:

$2{\left(\frac{|2x-y+4|}{\sqrt{5}}\right)}^2-3{\left(\frac{|x+2y-3|}{\sqrt{5}}\right)}^2=1\Rightarrow \frac{2}{5}{(2x-y+4)}^2-\frac{3}{5}{(x+2y-3)}^2=1$

Hence, the equation of the hyperbola is $\frac{2}{5}{(2x-y+4)}^2-\frac{3}{5}{(x+2y-3)}^2=1$.