If the pair of straight lines x^2 - 2pxy - y^2 = 0 and x^2 - 2qxy - y^2 = 0 be such that each pair bisects the angle between the other pair, then: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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If the pair of straight lines x^{2} - 2 p x y - y^{2} = 0 and x^{2} - 2 q x y - y^{2} = 0 be such that each pair bisects the angle between the other pair, then:

Solution & Step-by-step Explanation

The equation of the angle bisectors of a x^{2} + 2 h x y + b y^{2} = 0 is \frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h} .For x^{2} - 2 p x y - y^{2} = 0 : a = 1, h = - p, b = - 1 .Bisectors: \frac{x ^{2} - y ^{2}}{1 - ( - 1 )} = \frac{x y}{- p} ⟹ \frac{x ^{2} - y ^{2}}{2} = - \frac{x y}{p} ⟹ p (x^{2} - y^{2}) + 2 x y = 0 ⟹ x^{2} + \frac{2}{p} x y - y^{2} = 0 .Comparing with x^{2} - 2 q x y - y^{2} = 0 : - 2 q = \frac{2}{p} ⟹ - q = \frac{1}{p} ⟹ pq = - 1

Try it yourself before checking the explanation above.

If the pair of straight lines x^{2} - 2 p x y - y^{2} = 0 and x^{2} - 2 q x y - y^{2} = 0 be such that each pair bisects the angle between the other pair, then:

p = q

p = - q

pq = 1

pq = - 1

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