Question

If the foci of hyperbola lies on the line $y=x$, one asymptote is $y=2x$ and it is passing through the point $(3,4)$, then

• A. Equation of hyperbola is $3x^2-xy+2y^2=47$
• B. Equation of hyperbola is $2x^2-5xy+2y^2+10=0$
• C. Eccentricity of hyperbola is $\frac{\sqrt{17}}{4}$
• D. Eccentricity of hyperbola is $\frac{\sqrt{10}}{3}$

Answer 1

Answer: (B), (D)

The correct options are
B Equation of hyperbola is $2x^2-5xy+2y^2+10=0$
D Eccentricity of hyperbola is $\frac{\sqrt{10}}{3}$

Since one asymptote is $y=2x$ and transverse axis equation is $y=x$ then other asymptote is the image of $y=2x$ in the line $y=x$ i.e., $x=2y$
So, Hyperbola equation is $(x-2y)(2x-y)=k$
$\because$ It passes through $(3,4)\Rightarrow k=-10$
So, equation is $2x^2+2y^2-5xy+10=0$
Now let accute angle between asymptotes is $\theta$, then
$\tan \theta =\frac{2-\frac{1}{2}}{1+1}=\frac{3}{4}\Rightarrow \tan \theta /2=\frac{1}{3}[\because \theta \le 90]$
as we know that
$e^2={\sec }^2\theta /2=1+{\tan }^2\theta /2$
$\Rightarrow e=\frac{\sqrt{10}}{3}$