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

\frac{g}{g_{1}}

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\frac{g_{1}}{g}

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

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

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Solution

The correct option is B $\frac{g_{1}}{g}$
To get equation of pair of straight lines,

We have
$a x^{2} + 2 h x y + b y^{2} + 2 g x = 0$

$⟹ a x^{2} + 2 h x y + b y^{2} + 2 g x (1) = 0 - (1)$

$(a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} + 2 g_{1} x = 0$

$⟹ a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} = - 2 g_{1} x$

$⟹ \frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{- 2 g_{1} x} = 1$

Putting value of x in equation(1),
we get,
$a x^{2} + 2 h x y + b y^{2} + 2 g x (\frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{- 2 g_{1} x}) = 0$

$⟹ a x^{2} + 2 h x y + b y^{2} - g (\frac{a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2}}{g 1}) = 0$

$⟹ x^{2} (a - \frac{a_{1} g}{g_{1}}) + 2 (h - \frac{h_{1} g}{g_{1}}) x y + y^{2} (b - \frac{b_{1} g}{g_{1}}) = 0$

This is the equation of pair of straight lines and given that the lines are perpendicular,
hence Coefficient of $x^{2}$ +Coefficient of $y^{2}$ =0

$a - \frac{a_{1} g}{g_{1}} + b - \frac{b_{1} g}{g_{1}} = 0$

$⟹ \frac{a g_{1} - a_{1} g + b g_{1} - b_{1} g}{g_{1}} = 0$

$⟹ g_{1} (a + b) - g (a_{1} + b_{1}) = 0$

$⟹ \frac{(a 1 + b 1)}{a + b} = \frac{g 1}{g}$
(Hence proved)

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Similar questions

Q. Find the condition that the lines joining the origin to the points of intersection of the curves

a x^{2} + 2 h x y + b y^{2} 2 g x = 0

and

a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} + 2 g_{1} x = 0

be perpendicular.

Q. Show that the straight lines joining the origin to the other two points of intersection of the curves whose equations are

a x^{2} + 2 h x y + b y^{2} + 2 g x = 0

and

a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x = 0

will be at right angles if

g (a^{'} + b^{'}) - g^{'} (a + b) = 0 .

The line joining the origin to the points of intersection of the curves $a x^{2} + 2 h x y + b y^{2} + 2 g x = 0$ and $a^{^{'}} x^{2} + 2 h^{^{'}} x y + b^{^{'}} y^{2} + 2 g^{^{'}} x = 0$ will be mutually perpendicular, if

Q. The straight line joining the origin to the other two points of intersection of the curve whose equations are

a x^{2} + 2 h x y + 2 g x + b y^{2} = 0 a n d a^{'} x^{2} + 2 h^{'} x y + b^{'} y^{2} + 2 g^{'} x = 0

will be at right angle if

a_{1} x^{2} + 2 h_{1} x y + b_{1} y^{2} + 2 g_{1} x = 0, a_{2} x^{2} + 2 h_{2} x y + b_{2} y^{2} + 2 g_{2} x = 0

will be right angles is

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Pair of lines joining origin to the points of intersection of two curves ax2+2hxy+by2+2gx=0 and a1x2+2h1xy+b1y2+2g1x=0 will be at right angled to each other if a1+b1a+b=

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