Question

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

• A. $g(a^{\prime }+b^{\prime })-g^{\prime }(a+b)=0$
• B. $g{g}^{\prime }=a^{\prime }{b}^{\prime }-ab$
• C. $g-g^{\prime }=(a-b)(a^{\prime }-b^{\prime })$
• D. none of these

Answer 1

Answer: (A)

$g(a^{\prime }+b^{\prime })-g^{\prime }(a+b)=0$

Given eq of curves

$a{x}^2+2hxy+2gx+b{y}^2=0$----(1)

$a^{\prime }{x}^2+2h^{\prime }xy+2g^{\prime }x+b^{\prime }{y}^2=0$----(2)

$1=\frac{a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2}{-2g^{\prime }x}$

Putting in eq (1)

$a{x}^2+2hxy+2gx\left(\frac{a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2}{-2g^{\prime }x}\right)+b{y}^2=0$

$a{x}^2+2hxy-g\left(\frac{a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2}{g^{\prime }}\right)+b{y}^2=0$

$a{g}^{\prime }{x}^2+2h{g}^{\prime }xy-g{a}^{\prime }{x}^2-2g{h}^{\prime }xy-g{b}^{\prime }{y}^2+b{g}^{\prime }{y}^2=0$

$x^2(a{g}^{\prime }-a^{\prime }g)+y^2(b{g}^{\prime }-b^{\prime }g)+2xy(h{g}^{\prime }-g{h}^{\prime })=0$

Here by perpendicularlly

$a{g}^{\prime }-a^{\prime }g+b{g}^{\prime }-b^{\prime }g=0$

$(a+b)g^{\prime }-g(a^{\prime }+b^{\prime })=0$

$g(a^{\prime }+b^{\prime })-g^{\prime }(a+b)=0$