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  • If the curves  <img alt="\mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \text { and } A x^2+2 H x y+B y^2+2 G x+2 F y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathr

If the curves

 \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \text { and } A x^2+2 H x y+B y^2+2 G x+2 F y+c=0} inserts at 

four co cyclic  points then

 

 

Option: 1

\mathrm{\frac{(a+b)}{h}=\frac{(A+B)}{H}}


Option: 2

\mathrm{\frac{2(a+b)}{h}=\frac{(A+B)}{H}}


Option: 3

\mathrm{\frac{(a-b)}{h}=\frac{(A-B)}{H}}


Option: 4

None of these


Answers (1)

best_answer

Any second degree curve passing through the intersections of the given curves is


\mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c+\lambda\left(A X^2+2 H x y+B y^2+2 G x+2 F y+C\right)=0}

As the points of intersection of the two curves are co cyclic, (i) must be a circle for some λ.

\mathrm{\therefore \text { Coefficient of } \mathrm{x}^2=\text { coefficient of } \mathrm{y}^2 \text { and coefficient of } \mathrm{xy}=0 \text {. }}

\mathrm{\begin{aligned} & \therefore \mathrm{a}+\lambda \mathrm{A}=\mathrm{b}+\lambda \mathrm{B} \text { and } 2 \mathrm{~h}+\lambda \cdot 2 \mathrm{H}=0 \text { or } \mathrm{a}-\mathrm{b}=\lambda(\mathrm{B}-\mathrm{A}) \text { and } \mathrm{h}=-\lambda \mathrm{H} \\ & \therefore \mathrm{a}-\mathrm{b} / \mathrm{h}=\lambda(\mathrm{B}-\mathrm{A}) /-\lambda \mathrm{H} \quad \therefore(\mathrm{a}-\mathrm{b}) / \mathrm{h}=(\mathrm{A}-\mathrm{B}) / \mathrm{H} \end{aligned}}

Posted by

Sanket Gandhi

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