# Thread: locus

1. ## locus

given base of triangle and product of tangents of base angles then the locus of vertex of triangle is

2. Hi prasum,
Is it a right triangle?

bjh

3. 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.

4. i didnt understand your last step

5. 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 $\displaystyle \frac{y^2}{a(a- x)}= c$, 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.

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