Question

For hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$, and circle $x^2+y^2=100$, the ray from one of concyclic points passes through a focus.

• A. $10\sqrt{3}x+(10\sqrt{29}+41)y+20\sqrt{123}=0$
• B. $20\sqrt{3}x+(20\sqrt{29}+41)y+20\sqrt{123}=0$
• C. $20\sqrt{3}x+(10\sqrt{29}+41)y+20\sqrt{123}=0$
• D. $20\sqrt{3}x+(20\sqrt{29}+41)y+20\sqrt{41}=0$

Answer 1

Answer: (C)

$20\sqrt{3}x+(10\sqrt{29}+41)y+20\sqrt{123}=0$

The point of intersection is:

$y^2=100-x^2$ ...... $\left(1\right)$

And, $\frac{x^2}{25}-\frac{y^2}{16}=1$

$\frac{x^2}{25}-\frac{100-x^2}{16}=1$

$Or,\frac{{41x}^2}{25\ast 16}=\frac{116}{16}$

$Or,x=\pm \frac{10\sqrt{29}}{\sqrt{41}}$

And Using $\left(1\right)$ we get:

$y=\pm \frac{20\sqrt{3}}{\sqrt{41}}$

The point is $(\frac{10\sqrt{29}}{\sqrt{41}},-\frac{20\sqrt{3}}{\sqrt{41}})$

And we know, $e^2=1+\frac{16}{25}$

Or, $e=\frac{\sqrt{41}}{5}$

So, focus: $(-\sqrt{41},0)$

$\therefore$ Equation of line is:

$(y-0)=\frac{-\frac{20\sqrt{3}}{\sqrt{41}}-0}{+\frac{10\sqrt{29}}{\sqrt{41}}+\sqrt{41}}(x+\sqrt{41})$

$y=\frac{-20\sqrt{3}}{10\sqrt{29}+41}(x-\sqrt{41})$

$\Rightarrow (10\sqrt{29}+41)y=-20\sqrt{3}x-20\sqrt{123}$

$\Rightarrow 20\sqrt{3}x+(10\sqrt{29}+41)y+20\sqrt{123}=0$