Question

What's the equation of the line joining 2 points with eccentric angles ${30}^{\circ }$ and ${60}^{\circ }$ in the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$?

• A.
• B. (b)
• C. x~-~3y~-~15{\sqrt{3}}~+~8~=~0\)
• D. x~+~3y~+~15{\sqrt{3}}~-~8~=~0\)

Answer 1

The correct option is C

x~-~3y~-~15{\sqrt{3}}~+~8~=~0\)

Either we can take these 2 points and find the equation of the line or

we can use the formula for the same

Let the points be A and B

A $\equiv (4.sec{30}^{\circ },3.tan{30}^{\circ })=(\frac{8}{\sqrt{3}},\sqrt{3})$

B $\equiv (4.sec{60}^{\circ },3.tan{60}^{\circ })=(8,3\sqrt{3})$

The line adjoining these points ,is,

[x-8] $=\left[\frac{3\sqrt{3}-\sqrt{3}}{8-\frac{8}{\sqrt{3}}}\right].[y-3\sqrt{3}]$

$=\left[\frac{9-3}{8\sqrt{3}-8}\right]$ .$(y-3\sqrt{3})$

$4\sqrt{3}x-x-24\sqrt{3}+8=3y-9\sqrt{3}$

$(4\sqrt{3}-1)x-3y-15\sqrt{3}+8=0$

Method

we get the same answer by using the formula for line connecting 2 points $\left(\alpha \right)$ and $\left(\beta \right)$.

$\frac{x}{8}.cos\left(\frac{\propto -\beta }{2}\right)-\frac{y}{b}.sin\frac{\propto +\beta }{2}=cos\frac{\propto +\beta }{2}$