Question

If pair of straight lines $x^2-y^2+6x+4y+5=0$ are transverse and conjugate axes of hyperbola and perpendicular distance form origin to these lines represent the length of transverse and conjugate axis, then the eccentricity of hyperbola is:

• A. $\sqrt{3}$
• B. $\frac{\sqrt{26}}{5}$
• C. $\sqrt{26}$
• D. $2\sqrt{13}$

Answer 1

Answer: (B), (C)

$\sqrt{26}$

$x^2-y^2+6x+4y+5=0.....\left(1\right)$
Here $x^2-y^2=(x-y)(x+y)$
Seperate equation of lines are $(x-y+l)(x+y+m)$
In order to find the values of $l$ and $m$, we have to equation the coefficients of x and y
$x^2-xy+lx+xy-y^2+ly+mx-my+lm=0$
$\Rightarrow x^2-y^2+(l+m)x+(ly-my+lm)=0\cdots \left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right)$, we get
$l+m=6$ and $l-m=4$
$\therefore l=5$ and $m=1$
There are two possible cases for hyperbola equation i.e.,
$\frac{(x-y+5{)}^2}{(\sqrt{2}{)}^2{a}^2}-\frac{(x+y+1{)}^2}{b^2(\sqrt{2}{)}^2}=1\text{or}\frac{(x+y+1{)}^2}{(\sqrt{2}{)}^2{a}^2}-\frac{(x-y+5{)}^2}{(\sqrt{2}{)}^2{b}^2}=1$

Case $1$: If transverse axis equation is $x-y+5=0$ then length of transverse axis is
$2a=\left|\frac{0-0+5}{\sqrt{2}}\right|$
$\Rightarrow 2a=\frac{5}{\sqrt{2}}$
$\Rightarrow a=\frac{5}{2\sqrt{2}}$

If conjugate axis equation is $x+y+1=0$ then length of conjugate axis is
$2b=\left|\frac{0+0+1}{\sqrt{2}}\right|\Rightarrow b=\frac{1}{2\sqrt{2}}$
$\therefore e^2=1+\frac{b^2}{a^2}$
$\Rightarrow e=\frac{\sqrt{26}}{5}$

Case $2$: If transverse axis equation is $x+y+1=0$, then length of transverse axis is
$2a=\left|\frac{0+0+1}{\sqrt{2}}\right|\Rightarrow a=\frac{1}{2\sqrt{2}}$
Similarly $b=\frac{b}{2\sqrt{2}}$
$\therefore e^2=1+\frac{b^2}{a^2}$
$\Rightarrow e=\sqrt{26}$