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If $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$, then the angle between $x^2+2 x y \sec \theta+y^2=0$ is
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$\theta$
Given $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$
To find Angle between $x^2+2 x y \sec \theta+y^2=0$
Since, we know that angle between the line $a x^2+2 h x y+b y^2=0$ is
$\tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|$
Here, $\quad a=1$
$\tan \theta=\left|\frac{2 \sqrt{h^2-b}}{1+b}\right|$ ...(i)
For $x^2+2 x y \sec \theta+y^2=0$
$a=1, h=\sec \theta, b=1$
Let $\theta$ be the angle, then $\tan \phi=\left|\frac{2 \sqrt{\sec ^2 \theta-1}}{1+1}\right|$
$\tan \phi=\frac{2 \tan \theta}{2}=\tan \theta$
$\Rightarrow \quad \phi=\theta$
To find Angle between $x^2+2 x y \sec \theta+y^2=0$
Since, we know that angle between the line $a x^2+2 h x y+b y^2=0$ is
$\tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|$
Here, $\quad a=1$
$\tan \theta=\left|\frac{2 \sqrt{h^2-b}}{1+b}\right|$ ...(i)
For $x^2+2 x y \sec \theta+y^2=0$
$a=1, h=\sec \theta, b=1$
Let $\theta$ be the angle, then $\tan \phi=\left|\frac{2 \sqrt{\sec ^2 \theta-1}}{1+1}\right|$
$\tan \phi=\frac{2 \tan \theta}{2}=\tan \theta$
$\Rightarrow \quad \phi=\theta$
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