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Question: If the bisectors of the lines\(x^{2} - 2pxy - y^{2} = 0\)be \(x^{2} - 2qxy - y^{2} = 0,\) then...

If the bisectors of the linesx2−2pxy−y2=0x^{2} - 2pxy - y^{2} = 0be x2−2qxy−y2=0,x^{2} - 2qxy - y^{2} = 0, then

A

pq+1=0pq + 1 = 0

B

pq−1=0pq - 1 = 0

C

p+q=0p + q = 0

D

p−q=0p - q = 0

Answer

pq+1=0pq + 1 = 0

Explanation

Solution

Bisectors of the angle between the lines x2−2pxy−y2=0x^{2} - 2pxy - y^{2} = 0is x2−y2xy=1−(−1)−p\frac{x^{2} - y^{2}}{xy} = \frac{1 - ( - 1)}{- p}

⇒ px2+2xy−py2=0px^{2} + 2xy - py^{2} = 0

But it is represented by x2−2qxy−y2=0x^{2} - 2qxy - y^{2} = 0.

Therefore p1=2−2q⇒pq=−1⇒pq+1=0\frac{p}{1} = \frac{2}{- 2q} \Rightarrow pq = - 1 \Rightarrow pq + 1 = 0