Let $AB$ and $CD$ be two parallel chordes on hyperbola $y=1/x$. Lines $AC$ and $BD$ intersect axis $Oy$ at points $A_1$ and $D_1$ respectively, and axis $Ox$ - at points $C_1$ and $B_1$ respectively. Prove that the area of $\triangle A_1OC_1$ equals the area of $\triangle D_1OB_1$
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Tags: analytic geometry, geometry, conics, hyperbola
rafayaashary1
31.03.2016 16:35
chords? $ $ Edit: so im assuming that the points are ABCD in clockwise order, A at the "top"
proximo
31.03.2016 16:43
rafayaashary1 wrote: chords? $ $ Edit: so im assuming that the points are ABCD in clockwise order, A at the "top" I think it doesn't matter
RagvaloD
05.04.2016 02:11
For two points $X_1(x_1,1/x_1),X_2(x_2,1/x_2)$ on hyperbola equality of chord $X_1X_2$ is $$y=\frac{x_1+x_2-x}{x_1x_2}=-\frac{1}{x_1x_2}x+\frac{x_1+x_2}{x_1x_2}$$ Let $A,B,C,D$ have coordinates $(a,1/a),(b,1/b),(c,1/c)(d,1/d)$ $AB$ and $CD$ are parallel, so $\frac{1}{ab}=\frac{1}{cd}$ => $ab=cd$ $A_1,A,C,C_1$ lies on $y=\frac{a+c-x}{ac}$ Coordinates of $A_1$ are $(0,\frac{a+c}{ac})$, coordinates of $C_1$ are $(a+c,0)$. Area of $\triangle A_1OC_1$ is $\frac{(a+c)^2}{2ac}$ Similarly , area of $\triangle D_1OB_1$ is $\frac{(b+d)^2}{2bd}$ $\frac{(a+c)^2}{2ac}-\frac{(b+d)^2}{2bd}=\frac{a^2bd+2abcd+c^2bd-b^2ac-2abcd-d^2ac}{2abcd}=\frac{ab(ad-ac)+cd(cb-ad)}{2abcd}=\frac{(ab-cd)(cb-ad)}{2abcd}=0$ . So areas are equals.