Equations of Tangents and Normals

1. The Geometry of the Derivative: Tangents and Normals

At its heart, the derivative of a function at a point gives you the slope of the line that just "touches" the curve at that point without crossing it. This line is called the tangent. A second line, drawn perpendicular to the tangent at the same point, is called the normal.

The following video provides an excellent visual and conceptual introduction to tangents and normals.

From this, we can establish two fundamental facts:

1. The slope of the tangent to a curve $y=f(x)$ at a point $(x_1,y_1)$ is given by the value of the derivative at that point.

   $$m_{\text{tangent}}={\frac{dy}{dx}|}_{(x_1,y_1)}$$

2. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of the tangent's slope. (Recall from coordinate geometry that for two perpendicular lines with slopes $m_1$ and $m_2$, $m_1\cdot m_2=-1$).

   $$m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}=-\frac{1}{{\frac{dy}{dx}|}_{(x_1,y_1)}}$$

2. The Step-by-Step Process

Once you have the point $(x_1,y_1)$ and the slope ($m$), finding the equation of the line is a standard procedure from coordinate geometry using the point-slope form:

This is the primary tool you'll use to construct the final equations. The following resource provides a clear breakdown of the steps involved.

To summarize the complete process:

To find the equation of the tangent:

1. Differentiate: Find the derivative $\frac{dy}{dx}$ of the curve's equation.
2. Find Slope: Evaluate $\frac{dy}{dx}$ at the given point $(x_1,y_1)$ to get the slope, $m_{\text{tangent}}$.
3. Form Equation: Substitute $m_{\text{tangent}}$ and $(x_1,y_1)$ into the point-slope form: $y-y_1=m_{\text{tangent}}(x-x_1)$.

To find the equation of the normal:

1. Find Tangent Slope: Complete steps 1 and 2 from above to find $m_{\text{tangent}}$.
2. Find Normal Slope: Calculate $m_{\text{normal}}=-1/m_{\text{tangent}}$.
3. Form Equation: Substitute $m_{\text{normal}}$ and $(x_1,y_1)$ into the point-slope form: $y-y_1=m_{\text{normal}}(x-x_1)$.

3. Key Problem Variations for MHT-CET

For an exam like MHT-CET, problems often involve parametric equations or ask you to find the point of tangency itself. Let's explore these common variations.

The following video is an excellent resource from Pawan Wagh Academy, tailored specifically for MHT-CET preparation. It covers several important problem types.

Problem Type 1: Parametric Curves

This is a very frequently asked question type. A curve is defined by $x=f(t)$ and $y=g(t)$.

The process is:

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
2. Calculate the slope of the tangent: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.
3. Substitute the given value of the parameter (e.g., $t$ or $\theta$) into the expressions for $x$, $y$, and $\frac{dy}{dx}$ to find the point $(x_1,y_1)$ and the slope $m$.
4. Use the point-slope form.

Problem Type 2: Finding the Point on the Curve

Sometimes, you are given a condition on the tangent's slope (e.g., "the tangent is parallel to the line $y=2x+3$") and asked to find the point of tangency $(x,y)$.

The process is:

1. Find the required slope from the given condition. If the tangent is parallel to a line, the slopes are equal. If it's perpendicular, the slope is the negative reciprocal.
2. Find the derivative $\frac{dy}{dx}$ of the curve.
3. Set the derivative equal to the required slope from step 1: $\frac{dy}{dx}=m$.
4. Solve this equation for $x$.
5. Substitute the value(s) of $x$ back into the original curve's equation to find the corresponding $y$ value(s).

Problem Type 3 (Challenge): Finding Unknown Parameters

In a more advanced problem, you might be given the equation of the tangent and a curve with unknown constants (e.g., $y^2=p{x}^3+q$) and be asked to find those constants.

4. Practice Problems

Now, apply what you've learned to solve a few MHT-CET style questions.

Conclusion

In this lesson, we established a crucial link between calculus and geometry. You've learned how the derivative provides the slope of a tangent line, and from there, how to derive the equations for both the tangent and the perpendicular normal line.

Key Takeaways:

• The slope of the tangent at $(x_1,y_1)$ is $m_{\text{tangent}}={\frac{dy}{dx}|}_{(x_1,y_1)}$.
• The slope of the normal is $m_{\text{normal}}=-1/m_{\text{tangent}}$.
• The final equation for both lines is found using the point-slope form: $y-y_1=m(x-x_1)$.
• Be prepared for different problem formats, especially those involving parametric equations or requiring you to find the point of tangency.

We have now seen our first major application of derivatives. In the next lesson, we'll explore another powerful application: solving rate-of-change problems involving related rates. This will connect directly to your physics background, where you study how quantities like distance, velocity, and volume change with respect to time.