Question

The equation of the asymptotes of the hyperbola $2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-4=0$ are,
A. $2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-5=0$
B. $2{{x}^{2}}+4xy+2{{y}^{2}}-7x-11y+5=0$
C. $2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y+5=0$
D. None of these

Answer 1

Hint: The pair of the equation of asymptotes of the hyperbola is different from the equation of a hyperbola in constant term only. And the pair of equation of asymptotes represents the pair of straight-line if $\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ when $a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+2hxy+c=0$ is the pair of equation of straight line. We have the equation of hyperbola and using the above concept, we will find the value of c.

Complete step-by-step answer:

We are given that the equation of hyperbola is,
$2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-4=0...........\left( 1 \right)$

Now, we also know that the equation of pair of asymptotes is differ from only constant term that is equation of asymptotes is given by,
$2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y+\lambda =0...........\left( 2 \right)$
Now, we know that the equation of pair of asymptotes represents the equation of pair of straight line if,
$\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0.........\left( 3 \right)$
When the equation of pair of straight line given as,
$a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+2hxy+c=0........\left( 4 \right)$
By comparing equation (2) and (4), we get
$a=2,b=2,h=\dfrac{5}{2},g=\dfrac{-11}{2},f=\dfrac{-7}{2},c=\lambda $
So, now using equation (3),
$\begin{align}
  & \Delta =2\times 2\times \lambda +2\left( \dfrac{-7}{2} \right)\left( \dfrac{-11}{2} \right)\left( \dfrac{5}{2} \right)-2{{\left( \dfrac{-7}{2} \right)}^{2}}-2{{\left( \dfrac{-11}{2} \right)}^{2}}-\lambda {{\left( \dfrac{5}{2} \right)}^{2}}=0 \\
 & \Delta =4\lambda +77\times \dfrac{5}{4}-2\times \dfrac{49}{4}-2\times \dfrac{121}{4}-\dfrac{25}{4}\lambda =0 \\
 & \Rightarrow 4\lambda -\dfrac{25}{4}\lambda =\dfrac{49}{2}+\dfrac{121}{2}-\dfrac{385}{4} \\
 & \Rightarrow \dfrac{16\lambda -25\lambda }{4}=\dfrac{98+242-385}{4} \\
 & \Rightarrow -9\lambda =-45 \\
 & \Rightarrow \lambda =5 \\
\end{align}$
So, by putting the value of $\lambda $ in equation (2), we will get the equation of asymptotes as,
$2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y+5=0$
Hence, the correct option is (c).

Note: The possibility may have for the mistake is that students may choose the wrong option in a hurry to finish the question. Options A and C are very similar, so we have to check the sign of the last term carefully. Another possibility is that students may forget that the pair of asymptotes should represent the pair of a straight line which is why $\Delta $ must be equal to zero.