If on a given base triangles be described such that the sum of the tangents of the base angles is constant, prove that the locus of the vertices is a parabola.

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Solution

Let the base of triangle be of fixed length $2 a$ with its mid point on the origin

and the variable vertex be $A (h, k)$

In $△ A B D, tan ϕ_{3} = \frac{k}{h - a}$

In $△ A C D, tan ϕ_{1} = \frac{k}{h + a}$

$ϕ_{2} + ϕ_{3} = π ϕ_{2} = π - ϕ_{3}$

Given $tan ϕ_{1} + tan ϕ_{2} = c$

$tan ϕ_{1} + tan (π - ϕ_{3}) = c tan ϕ_{1} - tan ϕ_{3} = c \frac{k}{h + a} - \frac{k}{h - a} = c \frac{k h - k a - k h - k a}{h^{2} - a^{2}} = c h^{2} - a^{2} = - \frac{2 k a}{c} h^{2} = - \frac{2 k a}{c} + a^{2} h^{2} = - \frac{2 a}{c} (k - \frac{a c}{2})$

Replacing $h$ by $x$ and $k$ by $y$

$x^{2} = - \frac{2 a}{c} (y - \frac{a c}{2})$

which is a parabola

Hence proved.

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Describe the locus for questions 1 to 13 given below:

The locus of vertices of all isosceles triangles having a common base.

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