Question

The equation of a straight line, which passes through the point $(a{\cos }^3\theta ,a{\sin }^3\theta )$ and perpendicular to the line $xsec\theta +y\cos ec\theta =a$, is

• A. $x\cos \theta –y\sin \theta =a\cos 2\theta$
• B. $x\cos \theta +y\sin \theta =a\cos 2\theta$
• C. $x\cos \theta +y\sin \theta –a\cos 2\theta =1$
• D. $x\cos \theta -y\sin \theta +a\cos 2\theta =-1$

Answer 1

The correct option is A

$x\cos \theta –y\sin \theta =a\cos 2\theta$

Explanation For The Correct Option:

Finding the equation of straight line

The given point through which the line passes\

$(a{\cos }^3\theta ,a{\sin }^3\theta )$

And the given perpendicular line

$xsec\theta +y\cos ec\theta =a$

$\Rightarrow y=-\left(\frac{sec\theta }{\cos ec\theta }\right)x+a\sin \theta [Dividingboghsiesby\cos ec\theta ]$

Then we have the slope of this line,

$$\begin{gathered} \Rightarrow m_1=-\left(\frac{sec\theta }{\cos ec\theta }\right) \\ \Rightarrow m_1=-\left(\frac{\sin \theta }{\cos \theta }\right)\left[\because sec\theta \frac{1}{\cos \theta }\&\cos ec\theta =\frac{1}{\sin \theta }\right] \end{gathered}$$

Since the required line is perpendicular to this line so the slope of the line is,

$$\begin{gathered} \Rightarrow m_2=-\frac{1}{m_1} \\ \Rightarrow m_2=\frac{\cos \theta }{\sin \theta } \end{gathered}$$

And the line passes through the point $(a{\cos }^3\theta ,a{\sin }^3\theta )$

Since we know the equation of line having slope $m_2$, passing through $\left(x_1,y_1\right)$

$\Rightarrow y–y_1=m_2(x–x_1)$

Therefore, the equation of line be

$$\begin{gathered} \Rightarrow (y–a{\sin }^3\theta )=\left(\frac{\cos \theta }{\sin \theta }\right)(x–a{\cos }^3\theta ) \\ \Rightarrow y\sin \theta –a{\sin }^4\theta =x\cos \theta –a{\cos }^4\theta \\ \Rightarrow x\cos \theta –y\sin \theta =a{\cos }^4\theta –a{\sin }^4\theta \\ \Rightarrow x\cos \theta –y\sin \theta =a\left({\left({\cos }^2\theta \right)}^2–{\left({\sin }^2\theta \right)}^2\right) \\ \Rightarrow x\cos \theta –y\sin \theta =a({\cos }^2\theta +{\sin }^2\theta )({\cos }^2\theta –{\sin }^2\theta ) \\ \Rightarrow x\cos \theta –y\sin \theta =a\left(1\right)({\cos }^2\theta –{\sin }^2\theta ) \\ \Rightarrow x\cos \theta –y\sin \theta =a\cos 2\theta \end{gathered}$$

Hence option A is the correct answer.