Question:medium

Find the equation of the normal to a parabola which is perpendicular to a given line. This involves:

Show Hint

The slope of a normal to $y^2 = 4ax$ is often represented as $m$. The equation in slope form is $y = mx - 2am - am^3$. If you know the slope from the "given line," you can plug it directly into this formula!
Updated On: Apr 16, 2026
  • Slope comparison
  • Differentiation
  • Both A and B
  • None of these
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
To find the equation of a normal to a curve (like a parabola) that fulfills a specific geometric condition (being perpendicular to a given line), we must synthesize calculus and coordinate geometry techniques.
Step 2: Key Formula or Approach:
The slope of a tangent to a curve is found using differentiation $m_t = \frac{dy}{dx}$. The slope of the normal is $m_n = -1 / m_t$. Comparing slopes $m_1 \cdot m_2 = -1$ links the given line to the normal.
Step 3: Detailed Explanation:
Let's break down the procedure:
Slope comparison: Extract the slope $m_L$ of the given reference line. Since the normal is perpendicular to the given line, its slope $m_N$ must satisfy the perpendicularity condition $m_N \cdot m_L = -1 \implies m_N = -\frac{1}{m_L}$.
Differentiation: For a curve $y = f(x)$, the slope of the tangent is $\frac{dy}{dx}$. The slope of the normal is therefore $m_N = -\frac{1}{dy/dx}$. We set $-\frac{1}{dy/dx} = m_N$ to find the exact point of contact $(x_1, y_1)$ on the parabola.
Once the point is found, the point-slope form $y - y_1 = m_N(x - x_1)$ generates the final equation. Thus, both differentiation and slope comparison are indispensable steps in this process.
Step 4: Final Answer:
The process involves both Slope comparison and Differentiation.
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