AP EAMCET · Maths · Pair of Lines
If the equation of the pair of lines passing through \((1,1)\) and perpendicular to the pair of lines \(2 x^2+x y-y^2-x+2 y-1=0\) is \(a x^2+2 h x y+b y^2+2 g x+3 y=0\), then \(\frac{b}{a}=\)
- A \(\mathrm{g} / \mathrm{h}\)
- B \(2(g+h)\)
- C \(2(g-h)\)
- D gh
Answer & Solution
Correct Answer
(B) \(2(g+h)\)
Step-by-step Solution
Detailed explanation
\(2 x^2+x y-y^2-x+2 y-1=0 \implies A=2, B=1, C=-1\). Equation of perpendicular lines through \((1,1)\): \(C(x-1)^2 - B(x-1)(y-1) + A(y-1)^2 = 0\). \((-1)(x-1)^2 - (1)(x-1)(y-1) + (2)(y-1)^2 = 0\) \(-(x^2-2x+1) - (xy-x-y+1) + 2(y^2-2y+1) = 0\) \(-x^2+2x-1-xy+x+y-1+2y^2-4y+2 = 0\)…
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