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 Conic Triangle (Posted on 2014-10-17)

A line tangent to a hyperbola intersects its asymptotes at points L and N.
Let M be the intersection of the asymptotes.

Prove that the area of ΔLMN is a constant.

 Submitted by Bractals No Rating Solution: (Hide) The central equation for a hyperbola is``` x2 y2 ---- - ---- = 1 (1) a2 b2``` where 2a is the distance between the vertices and``` y = ±bx/a (2)``` are the equations of the asymptotes. Differentiating equation (1) gives``` b2x y' = ----- (3) a2y``` Because of the symmetry of the hyperbola, we only need to consider the point of tangency (xt,yt) in the first quadrant. The tangent line``` y = y'tx + k b2xt = ----- x + k a2yt b2xt2 yt = ------- + k a2yt b2xt b2xt2 y = ----- x + yt - ------ a2yt a2yt a2yty = b2xtx + a2yt2 - b2xt2 a2yty = b2xtx - a2b2 b2xt b2 y = ----- x - --- (4) a2yt yt``` Equation (4) and y = bx/a intersect at point``` L = (xl,yl) = ( a2b/[bxt - ayt] , ab2/[bxt - ayt] ) (5)``` which lies above the x-axis. Equation (4) and y = -bx/a intersect at point``` N = (xn,yn) = ( a2b/[bxt + ayt] , -ab2/[bxt + ayt] ) (6)``` which lies below the x-axis. Equation (4) and x-axis intersect at point``` I = (xi,yi) = ( a2/xt,0 ) (7)``` We can now calculate the area.``` Area(ΔLMN) = Area(ΔLMI) + Area(ΔNMI) = (a2/2xt)(ab2/[bxt - ayt] ) + (a2/2xt)(ab2/[bxt + ayt] ) = (a3b2/2xt)([bxt + ayt + bxt - ayt] /[b2xt2 - a2yt2 ]) = (a3b2/2xt)(2bxt/a2b2) = ab (8)``` Note: The above argument does not hold if the point of tangency is the vertex (a,0) because of equation (3). But, the area for this case is easily calculated.``` Area(ΔLMN) = (1/2)(base)(altitude) = (1/2)|LN||MI| = ab (9)``` Equations (8) and (9) agree on the constant ab. QED Note: See Ady's post for how simple the problem is for a rectangular hyperbola.

Comments: ( You must be logged in to post comments.)
 Subject Author Date re: Stretching a point? Bractals 2014-10-20 11:25:36 Stretching a point? Harry 2014-10-19 19:01:45 SOLUTION Ady TZIDON 2014-10-18 04:16:30
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