# locus

• Mar 19th 2011, 08:22 AM
prasum
locus
given base of triangle and product of tangents of base angles then the locus of vertex of triangle is
• Mar 19th 2011, 08:49 AM
bjhopper
Hi prasum,
Is it a right triangle?

bjh
• Mar 19th 2011, 09:47 AM
HallsofIvy
It take it you have a triangle ABC with the length of AB= a fixed and the product of the tangents of angles A and B fixed. Let x be the distance from A to the foot of the perpendicular from C. Then the distance from B to that point is a- x. If y is the distance from C to that foot, the two tangents are y/x and y/(a- x) so that constant product is y^2/a(a-x)= C. Solve for x as a function of y. The locus is the set of points, (x, y) satisfying that equation.
• Mar 19th 2011, 12:47 PM
prasum
i didnt understand your last step
• Mar 20th 2011, 03:49 AM
HallsofIvy
The "C" in the last equation is not, of course, the point, "C", but the product of tangents that you are given. It would have been better to call it, say, "c".

Solve $\frac{y^2}{a(a- x)}= c$, either for y as a function of x, or for x as a function of y (I recommended the latter). The locus is the set of points, (x, y), given by that. It should be obvious from the function what kind of curve this is.