Question

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points A and B.Find points A and B.

Answer 1

General point on hyperbola, $(3sec\theta ,2tan\theta )$

Putting in circle's equation,

$(3sec\theta {)}^2+(2tan\theta {)}^2-8\left(3sec\theta \right)=0$

$9se{c}^2\theta +4ta{n}^2\theta -24sec\theta =0$

$9se{c}^2\theta +4(se{c}^2\theta -1)-24sec\theta =0$

$13se{c}^2\theta -24sec\theta -4=0$

$sec\theta =\frac{24\pm \sqrt{576+208}}{26}$

$=\frac{24\pm \sqrt{784}}{26}$

$=\frac{24\pm 28}{26}=\frac{52}{26}$ or $\frac{-4}{26}=2$ or $\frac{-2}{13}$

$\Rightarrow tan\theta =\sqrt[\pm ]{\pm se{c}^2\theta -1}=\sqrt{4-1}=\pm \sqrt{3}$ or $\sqrt[\pm ]{\pm \frac{4}{169}-1}$

$\Rightarrow sec\theta =2,tan\theta =\sqrt[\pm ]{\pm 3}(<0)$

$\Rightarrow$ point A = $(6,2\sqrt{3})$

$\Rightarrow$ point B = $(6,-2\sqrt{3})$