JEE Main 2020 Application Of Derivatives — Let Normal Point Curve Intersect Axis

Quick Summary

Concept Tested: Implicit differentiation & normal‑tangent relationship | Chapter: Application of Derivatives | Difficulty: ★★★☆☆ | Estimated Time: 5 min | Key Formula: $$\frac{dy}{dx}=\frac{6x}{2y+1}$$ | Answer: 4 | Why: The tangent slope at the required point has magnitude $4$, so $|m|=4$.

The Question

For the curve $$y^{2}-3x^{2}+y+10=0$$, the normal at a point $P$ meets the $y$‑axis at $\left(0,\frac{3}{2}\right)$. Find $|m|$, where $m$ is the slope of the tangent at $P$.

Quick Answer

4. By implicit differentiation the tangent slope is $m=\dfrac{6x_{1}}{2y_{1}+1}$. Using the condition that the normal passes through $(0,\tfrac{3}{2})$ gives $y_{1}=1$ and $x_{1}=±2$, leading to $|m|=4$.

Video Solution

Understanding the Concept

When a curve is given implicitly, we differentiate the equation with respect to $x$ treating $y$ as a function of $x$. This yields the slope of the tangent $m=\dfrac{dy}{dx}$. The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: $m_{\text{normal}}=-\dfrac{1}{m}$. The point‑slope form of a line allows us to use a known point on the normal to determine the coordinates of the point of tangency.

Detailed Step-by-Step Solution

Step 1: Implicit Differentiation

Differentiate $$y^{2}-3x^{2}+y+10=0$$ with respect to $x$:

$$2y\frac{dy}{dx}-6x+\frac{dy}{dx}=0$$

Factor $\dfrac{dy}{dx}$:

$$\frac{dy}{dx}(2y+1)=6x\;\;\Longrightarrow\;\;\frac{dy}{dx}=\frac{6x}{2y+1}$$

Step 2: Slope of the Tangent at $P(x_{1},y_{1})$

Evaluate the derivative at the point $P$:

$$m=\left.\frac{dy}{dx}\right|_{P}=\frac{6x_{1}}{2y_{1}+1}$$

Step 3: Slope of the Normal

The normal is perpendicular to the tangent, hence

$$m_{\text{normal}}=-\frac{1}{m}=-\frac{2y_{1}+1}{6x_{1}}$$

Step 4: Equation of the Normal Using the Given Point

Using point‑slope form with the normal passing through $\left(0,\frac{3}{2}\right)$:

$$y-y_{1}=-\frac{2y_{1}+1}{6x_{1}}(x-x_{1})$$

Substituting $x=0,\;y=\frac{3}{2}$ gives

$$\frac{3}{2}-y_{1}=\frac{2y_{1}+1}{6}$$

Step 5: Solve for $y_{1}$

Clear the fraction by multiplying by $6$:

$$9-6y_{1}=2y_{1}+1$$
$$8=8y_{1}\;\;\Longrightarrow\;\;y_{1}=1$$

Step 6: Solve for $x_{1}$ Using the Original Curve

Insert $y_{1}=1$ into the original equation:

$$1^{2}-3x_{1}^{2}+1+10=0\;\;\Longrightarrow\;\; -3x_{1}^{2}+12=0$$
$$3x_{1}^{2}=12\;\;\Longrightarrow\;\;x_{1}^{2}=4\;\;\Longrightarrow\;\;x_{1}=±2$$

Case 1: $x_{1}=2$

For $x_{1}=2$, the tangent slope is

$$m=\frac{6(2)}{2(1)+1}=\frac{12}{3}=4$$

Case 2: $x_{1}=-2$

For $x_{1}=-2$, the tangent slope is

$$m=\frac{6(-2)}{2(1)+1}=\frac{-12}{3}=-4$$

Step 7: Compute the Required Absolute Value

Both cases give $|m|=4$, therefore

$$|m|=4$$

Final Answer

✅ $|m| = 4$

Essential Formulas for This Topic

$$\frac{dy}{dx}=\frac{-F_{x}}{F_{y}}\quad\text{for}\;F(x,y)=0$$
$$m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}$$
$$\text{Point‑slope form: }y-y_{0}=m(x-x_{0})$$

Note: $F_{x}$ and $F_{y}$ denote partial derivatives of the implicit function.

Common Mistakes to Avoid

Mistake 1: Using Tangent Slope Directly for the Normal

Some students write the normal line with slope $m$ instead of $-1/m$, which violates the perpendicularity condition.

Mistake 2: Forgetting to Substitute Back into the Original Curve

Finding $y_{1}$ alone is insufficient; $x_{1}$ must satisfy the original equation, otherwise the point $P$ may not lie on the curve.

Mistake 3: Ignoring the Absolute Value Requirement

Reporting $m=4$ or $m=-4$ without taking $|m|$ leads to a wrong final answer when the question explicitly asks for the magnitude.

Key Concept Summary

• Implicit differentiation provides $\dfrac{dy}{dx}$ for curves not given as $y=f(x)$.
• The normal’s slope is the negative reciprocal of the tangent’s slope.
• Use a known point on the normal to create an equation that determines the coordinates of the point of tangency.
• Always verify the obtained point satisfies the original curve equation.
• When the question asks for $|m|$, take the absolute value of the tangent slope.

Golden Rule: For any curve $F(x,y)=0$, the slope of the tangent at $(x_1,y_1)$ is $-\dfrac{F_x}{F_y}$ evaluated at that point.

Frequently Asked Questions

Q: Why do we need to differentiate implicitly?

A: The curve is given by an equation involving both $x$ and $y$. Implicit differentiation lets us find $\dfrac{dy}{dx}$ without solving for $y$ explicitly.

Q: How do we know the normal passes through $(0,\frac{3}{2})$?

A: It is stated in the problem. This extra point provides the necessary condition to determine $y_1$ (and subsequently $x_1$).

Q: What if $2y+1=0$ at some point?

A: The derivative formula $\dfrac{6x}{2y+1}$ would be undefined, indicating a vertical tangent. In this problem $y_1=1$, so the denominator is $3$, avoiding that case.

Q: Why are there two possible $x$‑coordinates?

A: Solving $x_{1}^{2}=4$ yields $x_{1}=±2$. Both points lie on the curve and satisfy the normal condition, leading to tangent slopes $4$ and $-4$, whose absolute value is the same.

Prerequisites to Solve This Question

1. Understanding of implicit differentiation.
2. Knowledge of the relationship between slopes of perpendicular lines.
3. Familiarity with the point‑slope form of a straight line.
4. Ability to solve simultaneous equations (substituting back into the original curve).

After Solving This, You Can:

• ✔ Apply implicit differentiation to any curve.
• ✔ Determine tangent and normal equations for points on implicit curves.
• ✔ Use given points on a normal (or tangent) to locate the point of tangency.
• ✔ Evaluate absolute values of slopes when required.

Study Tips for This Topic

1. Practice differentiating a variety of implicit equations to become comfortable with algebraic manipulation.
2. Memorize the perpendicular slope rule: $m_1 m_2 = -1$.
3. Always check that the point you find satisfies the original curve; the derivative alone does not guarantee this.
4. In JEE problems, the answer often asks for a magnitude or a specific numeric value—pay attention to absolute value signs.

Difficulty Rating & Exam Frequency

Difficulty: ★★★☆☆ (moderate)
Exam Frequency: Appears regularly in JEE Main and occasionally in JEE Advanced under the Calculus section.
Importance: High, as it tests core concepts of implicit differentiation and line geometry.

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Last Updated: July 2026
Question Source: JEE Main 2020 PYQ
Topic: Application Of Derivatives