Question

The normal to the curve \[ x=a(1+\cos\theta), \quad y=a\sin\theta \] at \(\theta\) always passes through a fixed point, then the fixed point is

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

Hint:

For parametric curves, first find \(\frac{dy}{dx}\) using \(\frac{dy/d\theta}{dx/d\theta}\). Then write the normal equation and test the given fixed points.

Answer 1

The Correct Option is A

Step 1: Differentiate the parametric equations to find the tangent slope.
The curve is $x = a(1 + \cos\theta)$, $y = a\sin\theta$. Differentiating: $\frac{dx}{d\theta} = -a\sin\theta$, $\frac{dy}{d\theta} = a\cos\theta$. By the chain rule: \[\frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta\] This is the slope of the tangent at parameter $\theta$.
Step 2: Find the slope of the normal.
The normal is perpendicular to the tangent. If tangent slope $= -\cot\theta$, then normal slope $= \tan\theta$ (negative reciprocal).
Step 3: Write the equation of the normal at the point.
The point on the curve is $(a(1+\cos\theta),\, a\sin\theta)$. Using point-slope form: \[y - a\sin\theta = \tan\theta \cdot \bigl(x - a(1+\cos\theta)\bigr)\]
Step 4: Test whether (a, 0) lies on every normal.
Substitute $x = a$ and $y = 0$: \[0 - a\sin\theta = \tan\theta \cdot \bigl(a - a(1+\cos\theta)\bigr)\] Right side: $\tan\theta \cdot (-a\cos\theta) = -\frac{\sin\theta}{\cos\theta} \cdot a\cos\theta = -a\sin\theta$.
Step 5: Verify the equality holds for all values of theta.
Left side $= -a\sin\theta$. Right side $= -a\sin\theta$. The equation holds for all $\theta$, confirming $(a, 0)$ lies on every normal.
Step 6: State the fixed point.
The fixed point through which every normal passes is \[\boxed{(a,\, 0)}\]