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Question

The straight line joining the origin to the other two points of intersection of the curve whose equations are $$\displaystyle ax^{2}+2hxy+2gx+by^{2}=0\: \: and\: \: a'x^{2}+2h'xy+b'y^{2}+2g'x=0$$ will be at right angle if


A
g(a+b)g(a+b)=0
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B
gg' = a'b' - ab
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C
g - g' = (a - b) (a' - b')
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D
none of these
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Solution

The correct option is B $$\displaystyle g\left ( a'+b' \right )-g'\left ( a+b \right )=0 $$
Given eq of curves 
$$ax^2+2hxy+2gx+by^2=0$$----(1)
$$a'x^2+2h'xy+2g'x+b'y^2=0$$----(2)
$$1=\dfrac{a'x^2+2h'xy+b'y^2}{-2g'x}$$
Putting in eq (1)
$$ax^2+2hxy+2gx\left ( \dfrac{a'x^2+2h'xy+b'y^2}{-2g'x} \right )+by^2=0$$
$$ax^2+2hxy-g\left ( \dfrac{a'x^2+2h'xy+b'y^2}{g'} \right )+by^2=0$$
$$ag'x^2+2hg'xy-ga'x^2-2gh'xy-gb'y^2+bg'y^2=0$$
$$x^2(ag'-a'g)+y^2(bg'-b'g)+2xy(hg'-gh')=0$$
Here by perpendicularity
$$ag'-a'g+bg'-b'g=0$$
$$(a+b)g'-g(a'+b')=0$$
$$g(a'+b')-g'(a+b)=0$$


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