Pair of lines joining origin to the points of intersection of two curves $a x^{2} + 2 h x y + b y^{2} + 2 g x = 0 a n d a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} + 2 g_{1} x = 0$ will be at right angled to each other if $\frac{a_{1} + b_{1}}{a + b} =$

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$\frac{g}{g_{1}}$

$\frac{g_{1}}{g}$

$\frac{- g}{g_{1}}$

$\frac{- g_{1}}{g}$

answer is B.

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Detailed Solution

Given Equation

$\begin{array}{l} a x^{2} + 2 h x y + b y^{2} + 2 g x = 0 - - (1) a n d \\ a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} = - 2 g_{1} x \\ \Rightarrow \frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{- 2 g_{1} x} = 1 - - (2) \\ H o m o g e n i z i n g (1) w i t h (2) \\ a x^{2} + 2 h x y + b y^{2} + 2 g x (1) = 0 \\ \Rightarrow a x^{2} + 2 h x y + b y^{2} + 2 g x (\frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{- 2 g_{1} x}) = 0 \\ \Rightarrow a x^{2} + 2 h x y + b y^{2} + g (\frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{- g_{1}}) = 0 \\ T h e l i n e s a r e ⊥ r i f f \end{array}$

$\begin{array}{l} c o . e f f x^{2} + c o . e f f y^{2} = 0 \\ \Rightarrow a + b - \frac{g a_{1}}{g_{1}} - \frac{g b_{1}}{g_{1}} = 0 \\ \Rightarrow (a + b) g_{1} - g (a_{1} + b_{1}) = 0 \\ \Rightarrow (a_{1} + b_{1}) g = (a + b) g_{1} \\ \Rightarrow \frac{a_{1} + b_{1}}{a + b} = \frac{g_{1}}{g} \end{array}$