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  • The straight lines joining the origin to the points of intersection of the curves <img alt="\mathrm{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}" src="https://entrancec

The straight lines joining the origin to the points of intersection of the curves \mathrm{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 angles if
 

Option: 1

\mathrm{g}(\mathrm{a}+\mathrm{b})=\mathrm{g} 1(\mathrm{a} 1+\mathrm{b} 1)

 


Option: 2

\mathrm{g(a 1+b 1)=g 1(a+b)}
 


Option: 3

\mathrm{\operatorname{gg} 1=(\mathrm{a}+\mathrm{b})(\mathrm{a} 1+\mathrm{b} 1)}
 


Option: 4

\mathrm{hh1 =\mathrm{gg} 1}


Answers (1)

best_answer

The two curves are

\mathrm{ a x^2+2 h x y+b y^2+2 g x=0 }     ......(i)

\mathrm{ and \: a_1 x^2+2 h_1 x y+b_1 y^2+2 g_1 x=0 }  ......(ii)

Eliminating first degree terms in (i) and (ii) by multiplying (i) by g1 and (ii) by g and then subtracting the equation of the lines joining the points of intersection of (i) and (ii) to the origin is

\mathrm{ \mathrm{g}_1\left(\mathrm{ax}^2+2 \mathrm{hxy}+b \mathrm{y}^2\right)-\mathrm{g}\left(\mathrm{a}_1 \mathrm{x}^2+2 \mathrm{~h}_1 \mathrm{xy}+\mathrm{b}_1 \mathrm{y}^2\right)=0 }

\mathrm{ \Rightarrow \quad\left(\mathrm{ag}_1-\mathrm{a}_1 \mathrm{~g}\right) \mathrm{x}^2+2\left(\mathrm{hg}_1-\mathrm{h}_1 \mathrm{~g}\right) \mathrm{xy}+\left(\mathrm{bg}_1-\mathrm{b}_1 \mathrm{~g}\right) \mathrm{y}^2=0 }

These lines are mutually perpendicular, if \mathrm{ \text { coeff. of } \mathrm{x}^2+\text { coeff. of } \mathrm{y}^2=0 }

\mathrm{ \text { or }\left(\mathrm{ag}_1-\mathrm{a}_1 \mathrm{~g}\right)+\left(\mathrm{bg}_1-\mathrm{b}_1 \mathrm{~g}\right)=0 }

\mathrm{ \text { or } \mathrm{g}_1(\mathrm{a}+\mathrm{b})=\mathrm{g}\left(\mathrm{a}_1+\mathrm{b}_1\right) }

Hence option 2 is correct.



 

Posted by

Pankaj Sanodiya

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