Question

Find the normal to the curve $x=a\left(1+\cos \left(\theta \right)\right)$, $y=a\sin \left(\theta \right)$ at $\theta$ always passes through

• A. $\left(a,a\right)$
• B. $\left(0,a\right)$
• C. $\left(0,0\right)$
• D. $\left(a,0\right)$

Answer 1

The correct option is D

$\left(a,0\right)$

Explanation for the correct option

Step 1: Solve for the slope of the normal

Given equation of the curve is $x=a\left(1+\cos \left(\theta \right)\right)$, $y=a\sin \left(\theta \right)$

We know that the slope of the tangent to a curve $=\frac{dy}{dx}$

So the slope of the tangent to the curve is,
$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}\\ & =& \frac{{\displaystyle \frac{d}{d\theta }}\left(a\sin \left(\theta \right)\right)}{{\displaystyle \frac{d}{d\theta }}\left[a\left(1+\cos \left(\theta \right)\right)\right]}\\ & =& \frac{a\cos \left(\theta \right)}{a\left(-\sin \left(\theta \right)\right)}\\ & =& -\cot \left(\theta \right)\end{array}$

The slope of the normal $=-\frac{1}{{\displaystyle \frac{dy}{dx}}}$

Thus, the slope of the normal of the given curve is $-\frac{1}{\left(-\cot \left(\theta \right)\right)}=\tan \left(\theta \right)$

Step 2: Solve for the required point

The equation of the normal can be obtained from the point-slope form of line which is given as,
$y-y_1=m\left(x-x_1\right)$
where $m$ is the slope and $\left(x_1,y_1\right)$is the point through which line passes

Thus, the equation of the normal is,
$\begin{array}{ccc}& y-a\sin \left(\theta \right)=\tan \left(\theta \right)\left[x-a\left(1+\cos \left(\theta \right)\right)\right]& \\ \Rightarrow & y-a\sin \left(\theta \right)=\tan \left(\theta \right)\left[x-a-a\cos \left(\theta \right)\right]& \\ \Rightarrow & y-a\sin \left(\theta \right)=x\tan \left(\theta \right)-a\tan \left(\theta \right)-a\tan \left(\theta \right)·\cos \left(\theta \right)& \\ \Rightarrow & y-\cancel{a\sin \left(\theta \right)}=\tan \left(\theta \right)\left(x-a\right)-\cancel{a\sin \left(\theta \right)}& \left[\because \tan \left(\theta \right)\cos \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}·\cos \left(\theta \right)=\sin \left(\theta \right)\right]\\ \Rightarrow & y-0=\tan \left(\theta \right)\left(x-a\right)& \end{array}$

Comparing the above equation with the general equation of a line given earlier,
$x_1=a$ and $y_1=0$

Therefore, the normal of the given curve always passes through $\left(a,0\right)$

Hence, option(D) i.e. $\left(a,0\right)$ is correct.