Question

In an obtuse-angled triangle, the obtuse angle is $\frac{3\pi }{4}$​ and the other two angles are equal to two values of $\theta$ satisfying $a\tan \theta +b\sec \theta =c$, where $|b\mid \le a^2+c^2$​, then $a^2-c^2=kac$, then the distance from origin to the line $x+ky-2\sqrt{5}=0$ is

Answer 1

$a\tan \theta +b\sec \theta =c$
$\Rightarrow b^2{\sec }^2\theta =(c-a\tan \theta {)}^2$
$\Rightarrow b^2(1+{\tan }^2\theta )=c^2-2ca\tan \theta +a^2{\tan }^2\theta$
$\Rightarrow (a^2-b^2){\tan }^2\theta -2ac\tan \theta +c^2-b^2=0\cdots \left(i\right)$
Roots of equation $\left(i\right)$ are $\tan \alpha$ and $\tan \beta$, where $\alpha$ and $\beta$ are the two angles of the triangle.
We have, $\tan \alpha +\tan \beta =\frac{2ca}{a^2-b^2}$
and, $\tan \alpha .\tan \beta =\frac{c^2-b^2}{a^2-b^2}$
$\therefore \tan (\alpha +\beta )=\frac{\frac{2ca}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}=\frac{2ca}{a^2-c^2}$
$\therefore \tan (\pi -\frac{3\pi }{4})=\frac{2ca}{a^2-c^2}$
$\Rightarrow a^2-c^2=2ca$
$\therefore k=2$
Hence distance from origin to the line $x+ky-2\sqrt{5}=0$ is $\frac{2\sqrt{5}}{\sqrt{k^2+1}}=2$