Question

The points on the curve \( y^2 = x + \sin x \) at which the normal is parallel to the Y-axis lie on

• A. a line parallel to Y-axis
• B. a circle with centre at origin
• C. a parabola
• D. a pair of lines bisecting the angle between the coordinate axes

Hint:

The normal to a curve is parallel to the Y-axis when its slope is undefined, which means the slope of the tangent is zero. Find the derivative \( \frac{dy}{dx} \) and set \( \frac{dx}{dy} = 0 \). Use this condition along with the equation of the curve to find the relationship between \( x \) and \( y \) for the points where the normal is parallel to the Y-axis.

Answer 1

The Correct Option is C

The equation of the curve is \( y^2 = x + \sin x \).
Differentiating with respect to \( x \): $$ 2y \frac{dy}{dx} = 1 + \cos x $$ The slope of the tangent is \( m_T = \frac{dy}{dx} = \frac{1 + \cos x}{2y} \).
The slope of the normal is \( m_N = -\frac{1}{m_T} = -\frac{2y}{1 + \cos x} \).
If the normal is parallel to the Y-axis, its slope is undefined.
This means the denominator of \( m_N \) must be zero, and the numerator must be non-zero.
$$ 1 + \cos x = 0 \implies \cos x = -1 $$ This occurs when \( x = (2n + 1)\pi \), where \( n \) is an integer.
At these values of \( x \), the numerator of \( m_N \) is \( -2y \).
For the slope to be undefined, \( y \) must be non-zero.
Now, substitute \( \cos x = -1 \) into the equation of the curve \( y^2 = x + \sin x \): $$ y^2 = (2n + 1)\pi + \sin((2n + 1)\pi) $$ Since \( \sin((2n + 1)\pi) = 0 \), we have: $$ y^2 = (2n + 1)\pi $$ The points at which the normal is parallel to the Y-axis satisfy \( x = (2n + 1)\pi \) and \( y^2 = (2n + 1)\pi \).
Substituting \( x \) for \( (2n + 1)\pi \) in the second equation, we get \( y^2 = x \).
This is the equation of a parabola with its vertex at the origin and axis along the X-axis.
The points lie on the curve \( y^2 = x \), which is a parabola.