There are some methods to construct a conic, example: Based on Pascal theorem, Steiner construction, .....I propose a method to construct a conic as follows:

Let $L_1, L_2$ be two parallel lines, let $A, B, C, D$ be four points in the plane. Let $E$ be a point lie on the line $L_1$, $F$ be the point lie on line $L_2$ such that $EF \parallel AB$. Let circle $(E, ED)$ meets the circle $(F, FC)$ at two points $H$, $G$.

My question: I am looking for a proof that locus of $H, G$ is a conic section when $E$ be moved on line $L_1$.

enter image description here

See also:


Without loss of generality, let $L_1$ be the x-axis and $t$ be a parameter. Denote $C,D,E,F$ by $$C(x_1,y_1),D(x_2,y_2),E(t,0),F(t+a,b)$$ such that the vector $<a,b>$ is in the same direction as $\overrightarrow{AB}$ and $x_1,y_1,x_2,y_2,a,b$ are constants. Then the locus of $H$ (or $G$) satisfies $$\frac{PE}{ED}=\frac{PF}{FC}=1,$$ where $P(x,y)$ represents $H$ (or $G$). In terms of equations, one has $$(x-t)^2+y^2=(x_2-t)^2+y_2^2\qquad (1)$$ and $$(x-t-a)^2+(y-b)^2=(x_1-t-a)^2+(y_1-b)^2\qquad (2)$$ Then from (1), one gets $$t=\frac{x^2+y^2-x_2^2-y_2^2}{2(x-x_2)}\qquad (3)$$ Subtracting (2) from (1) gives $$(2x-2t-a)(a)+(2y-b)(b)=(x_1+x_2-2t-a)(x_2-x_1+a)+(y_2+y_1-b)(y_2-y_1+b)\quad (4)$$ Now substituting (3) into (4) and clearing denominator, one gets a quadratic equation in $x$ and $y$, so it gives an equation for a conic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.