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

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$$ag'-a'g+bg'-b'g=0$$$$(a+b)g'-g(a'+b')=0$$$$g(a'+b')-g'(a+b)=0$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More