Equation of the hyperbola passing through the point (1,−1) and having asymptotes x+2y+3=0 and 3x+4y+5=0 is :
A
3x2−10xy+8y2−14x+22y+15=0
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B
3x2+10xy+8y2−14x+22y+35=0
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C
3x2+10xy+8y2+14x+22y−8=0
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D
3x2+10xy+8y2+14x+22y+7=0
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Solution
The correct option is D3x2+10xy+8y2+14x+22y+7=0 Pair of Asymptotes: (x+2y+3)(3x+4y+5)=0 ⇒3x2+8y2+10xy+22y+14x+15=0 We know that equation of hyperbola and (Pair of asymptotes) differs by a constant. So required family of hyperbola is 3x2+8y2+10xy+22y+14x+15+λ=0 Since hyperbola passes through (1,−1) ⇒3(1)2+8(−1)2+10(1)(−1)+22(−1)+14(1)+15+λ=0 ⇒3+8+(−10)−22+14+15+λ=0 ⇒λ=−8
Putting the value of λ in Equation, we get required hyperbola equation as 3x2+10xy+8y2+14x+22y+7=0