Question

Prove that the equation to the straight line which passes through the point $(a{\cos }^3\theta ,a{\sin }^3\theta )$ and perpendicular to the straight line $xsec\theta +ycosec\theta =a$ is $x\cos \theta -y\sin \theta =a\cos 2\theta$.

Answer 1

The given line is $xsec\theta +ycosec\theta =a$

Slope is $m=\frac{-sec\theta }{cosec\theta }$

Then, the slope of perpendicular is $m_1=\frac{cosec\theta }{sec\theta }=cot\theta$

Then the equation of the line with this slope is $y=cot\theta x+c$.......$\left(1\right)$

As the line passes through the point $(aco{s}^3\theta ,asi{n}^3\theta )$

Then, $asi{n}^3\theta =cot\theta \times aco{s}^3\theta +c$

$\Rightarrow c=a\frac{si{n}^4\theta -co{s}^4\theta }{sin\theta }$

Therefore required line is $xcos\theta -ysin\theta =acos2\theta$ (Substitute ${}^{\prime }{c}^{\prime }$ in $\left(1\right)$ and simplify)