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Question: The equation of the asymptotes of the hyperbola \(2{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-4=0\) are, A. \...

The equation of the asymptotes of the hyperbola 2x2+5xy+2y2−11x−7y−4=02{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-4=0 are,
A. 2x2+5xy+2y2−11x−7y−5=02{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y-5=0
B. 2x2+4xy+2y2−7x−11y+5=02{{x}^{2}}+4xy+2{{y}^{2}}-7x-11y+5=0
C. 2x2+5xy+2y2−11x−7y+5=02{{x}^{2}}+5xy+2{{y}^{2}}-11x-7y+5=0
D. None of these

Explanation

Solution

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 Δ=abc+2fgh−af2−bg2−ch2=0\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 when ax2+by2+2gx+2fy+2hxy+c=0a{{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,
2x2+5xy+2y2−11x−7y−4=0...........(1)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,
2x2+5xy+2y2−11x−7y+λ=0...........(2)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,
Δ=abc+2fgh−af2−bg2−ch2=0.........(3)\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,
ax2+by2+2gx+2fy+2hxy+c=0........(4)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=52,g=−112,f=−72,c=λa=2,b=2,h=\dfrac{5}{2},g=\dfrac{-11}{2},f=\dfrac{-7}{2},c=\lambda
So, now using equation (3),
Δ=2×2×λ+2(−72)(−112)(52)−2(−72)2−2(−112)2−λ(52)2=0 Δ=4λ+77×54−2×494−2×1214−254λ=0 ⇒4λ−254λ=492+1212−3854 ⇒16λ−25λ4=98+242−3854 ⇒−9λ=−45 ⇒λ=5 \begin{aligned} & \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{aligned}
So, by putting the value of λ\lambda in equation (2), we will get the equation of asymptotes as,
2x2+5xy+2y2−11x−7y+5=02{{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 Δ\Delta must be equal to zero.