## Some solved examples based on pairs of straight lines

The following questions use the previous (previous blog)theory, concepts and formulae related to pair of straight lines:

Problem 1:

Prove that the straight lines joining the origin to the points of intersection of the straight line $hx + ky =2hk$ and the curve $(x-k)^{2}+(y-h)^{2}=c^{2}$ are at right angles if $h^{2}+k^{2}=c^{2}$.

Solution 1:

Making the equation of the curve homogeneous with the help of the equation of the line, we get

$x^{2}+y^{2} - 2(kx+hy)(\frac{hx+ky}{2hk})+(h^{2}+k^{2}-c^{2})(\frac{hx+ky}{2hk})^{2}=0$, or

$4h^{2}k^{2}x^{2} + 4h^{2}k^{2}y^{2}-4hk^{2}x(hx+ky) - 4h^{2}ky(hx+ky)+(h^{2}+k^{2}-c^{2})(h^{2}x^{2}+k^{2}y^{2}+2hxy)=0$

This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if

coefficient of $x^{2}$ + coefficient of $y^{2}$ is zero, that is,

$\Longrightarrow (h^{2}+k^{2})(h^{2}+k^{2}-c^{2})=0$

$\Longrightarrow h^{2}+k^{2}=c^{2}$ (since $h^{2}+k^{2} \neq 0$).

Problem 2:

Prove that the two straight lines $(x^{2}+y^{2})(\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta})=(x\tan{\alpha}+y\sin{\theta})^{2}$ include an angle $2\alpha$.

Solution 2:

The given equation can be written as

$(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta} - \tan^{2}{\alpha})x^{2} + 2\tan{\alpha}\sin{\theta}xy + \cos^{2}{\theta}\sin^{2}{\alpha}y^{2}=0$

Here, $a=\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha}$,

$b=\cos^{2}{\theta}\sin^{2}{\alpha}$ and $h=\tan{\alpha}\sin{\theta}$.

We need to show  that

$2\alpha = \arctan {2\frac{\sqrt{h^{2}-ab}}{a+b}}$

or $\tan{2\alpha} = \frac{2\sqrt{h^{2}-ab}}{a+b}$

Now, $\frac{2\sqrt{h^{2}-ab}}{a+b}$

$= \frac{2\sqrt{\tan^{2}{\alpha}\sin^{2}{\theta}-(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha})\cos^{2}{\theta}\sin^{2}{\alpha}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{\sin^{2}{\theta}-\cos^{4}{\theta}\sin^{2}{\alpha}\cos^{2}{\alpha}-\sin^{2}{\theta}\cos^{2}{\theta}\cos^{2}{\alpha}+\sin^{2}{\alpha}\cos^{2}{\theta}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+1-\cos^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{1-\cos^{2}{\theta}-\cos^{4}{\theta}{\cos^{2}{\alpha(1-\cos^{2}{\alpha})-(1-\cos^{2}{\theta})\cos^{2}{\theta}\cos^{2}{\alpha}+(1-\cos^{2}{\alpha})\cos^{2}{\theta}}}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(1-2\sin^{2}{\alpha})}$

$=\frac{2\tan{\alpha}\sqrt{1+\cos^{4}{\theta}\cos^{4}{\alpha}-2\cos^{2}{\theta}\cos^{2}{\alpha}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(\cos^{2}{\theta}-\sin^{2}{\alpha})}$

$= \frac{2\tan{\alpha}(1-\cos^{2}{\theta}\cos^{2}{\alpha})}{(1-\tan^{2}{\alpha})(1-\cos^{2}{\theta}\cos^{2}{\alpha})}=\tan{2\alpha}$

Hence, the given lines include an angle $2\alpha$.

More later,

Nalin Pithwa.