Name: Conjugate Hyperbola
Uploaded: 2019-07-19
Duration: 12 min 19 s
Description: In this video, we will learn about the transverse axis and conjugate axis of any hyperbola with respect to another hyperbola. Watch and learn.

Question

Given the base of a triangle and the ratio of the tangents of half of the base of the angles, prove that the vertex moves on a hyperbola whose foci are extremities of the base.

Accepted Answer

Let $A B C$ be any triangle whose sides $B C$ , $C A$ and $A B$ are $a$ , $b$ and $c$ , respectively, and let $S$ be equal to $\frac{1}{2} (a + b + c)$ and $r$ be the radius of inner circle of the triangle $A B C$ . Then, we get,

$\tan \frac{A}{2} = \frac{r}{s - a}$

$\tan \frac{B}{2} = \frac{r}{s - b}$

$\tan \frac{c}{2} = \frac{r}{s - c}$

Let $A B$ be the base and $C$ be the moving point of the triangle. Then, $A$ and $B$ are the base angles of the triangle.

Hence, by hypothesis, we get,

$\frac{\tan (\frac{A}{2})}{\tan (\frac{B}{2})} = \frac{\frac{r}{(s - a)}}{\frac{r}{(s - b)}}$

$\frac{\tan (\frac{A}{2})}{\tan (\frac{B}{2})} = \frac{s - b}{s - a}$

$\frac{\tan (\frac{A}{2})}{\tan (\frac{B}{2})} = λ$

$λ = \frac{\frac{a + b + c}{2} - a}{\frac{a + b + c}{2} - b}$

$λ = \frac{b + c - a}{c + a - b}$

$\frac{λ}{1} = \frac{b + c - a}{c + a - b}$

Hence, by componendo and dividendo, we get,

$\frac{1 + λ}{1 - λ} = \frac{2 c}{2 (b - a)}$

$\frac{1 + λ}{1 - λ} = \frac{c}{b - a}$
$b - a = \frac{c (1 - λ)}{1 + λ}$

As $c$ and $λ$ are constant, $(b - a)$ is constant. Therefore, the locus of the vertex $C$ is a hyperbola.

Given the base of a triangle and the ratio of the tangents of half of the base of the angles, prove that the vertex moves on a hyperbola whose foci are extremities of the base.

Important Questions on The Hyperbola

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