Animated Solution for Mathematics - Differentiation: The normal to the curve $x=a(1+\cos \theta )$, $y=a\sin \theta$ at '$\theta$' always passes through the fixed point

The parametric equations are given by $x=a(1+\cos \theta )$ and $y=a\sin \theta$. As $\theta$ varies, these equations trace a specific path in the Cartesian plane.

Our objective is to determine a fixed point through which the normal line to this curve passes, regardless of the value of the parameter $\theta$.

To define the normal line, we first require the slope of the tangent, denoted by $\frac{dy}{dx}$. Using the chain rule for parametric differentiation:

Differentiating the parametric equations with respect to $\theta$:

Thus, the slope of the tangent is:

Since the normal is perpendicular to the tangent, its slope $m$ is the negative reciprocal of the tangent's slope:

Using the point-slope form $y-y_1=m(x-x_1)$ with point $(a(1+\cos \theta ),a\sin \theta )$ and slope $\tan \theta$:

Substituting $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and cross-multiplying by $\cos \theta$:

Expanding both sides yields:

Canceling the common term $-a\sin \theta \cos \theta$ from both sides, we obtain the simplified equation:

For this equation to hold for any $\theta$, we examine the relationship $y=(x-a)\tan \theta$. This equation is satisfied independently of $\theta$ if and only if the coefficient of $\tan \theta$ is zero and $y$ is zero.

Setting $x-a=0$ gives $x=a$, which subsequently forces $y=0$. Thus, the fixed point is $(a,0)$.

To verify this, we observe the original equations:

This confirms the curve is a circle centered at $(a,0)$ with radius $a$. Geometrically, the normal to any circle must always pass through its center, confirming our algebraic result.