Tangent and Normal line

---

---

---

The derivative of a function has many applications to problems in calculus. It is useful to solve problems on (a) rate of change (b) slope/equation of tangent and normal lines (c) increasing and decreasing points/intervals (d) natute of points (e) shape of graph (f) curve sketching (g) approximation of function (h) maximum and minimum problems (i) distance; velocity, and acceleration problems

In this section, we use derivative to solve problems on (b) slope/equation of tangent and normal lines. The derivative of a function at a point is the slope of the tangent line at this point. The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency.

A tangent is a straight line that touches a curve at one point. The idea is that the tangent and the curve are both going in exact same direction at point of contact. In a space curve, number of lines may pass touching one point, therefore, a precise definition of tangent is given as below.
Let C be a curve.
Also let A and B are two points on C
when, point B is moved toward A then limiting form of secant AB, is called tangent to the curve C at A.

Drag point A	Drag point B

Equation of tangent line

Let y=f(x) be a curve C
Also let \(P(x_1,y_1)\) be a given point on C.
Then
Equation of straight line passing through P is
\( y-y_1=m(x-x_1)\)
Since, the tangent line has slope \(m=\frac{dy}{dx}\), the equation of tangent line at P is
\( y-y_1=\frac{dy}{dx} (x-x_1)\)
or \( y=y_1+\frac{dy}{dx} (x-x_1)\)
This completes the proof

---

---

---

Examples : Activity 1

1. Find the slope of tangent to the curve \(y=x^3-x\) at x=2

   Solution 👉 Click Here

   Given equation of curve is
   \(y=x^3-x\) (1)
   Given point is
   \((x_1,y_1)=(x,f(x))=(2,f(2))=(2,6)\)
   The equation of tangent line at \((x_1,y_1)\) is
   \( y−y_1=\frac{𝑑𝑦}{𝑑𝑥}(x−x_1)\)
   or \(y−6= \frac{𝑑𝑦}{𝑑𝑥}(x−2)\)
   The slope of the curve at x=2 is
   \(\frac{𝑑𝑦}{𝑑𝑥}= \frac{𝑑(x3-x )}{dx} =3x^2-1\)
   or \( \left | \frac{𝑑𝑦}{𝑑𝑥} \right |_{𝑥=2}= \left |3x^2-1 \right |_{𝑥=2}=11 \)
   So, the equation of tangent line at x=2 is
   \( y−6=\frac{𝑑𝑦}{𝑑𝑥}(x−2)\)
   or \( y−6=11(x−2)\)
   

2. Find the equation of tangent line to
   1. to the parabola \(y=x^2\) at the point (1,1)
   2. to the hyperbola \(y=\frac{3}{x}\) at the point (3,1)
   3. to the function \(y=\sqrt{x}\) at the point (1,1), (4,2) and (9,3)
   4. to the parabola \(y=x^2+2x\) at (-3,3)
   5. to the parabola \(y=x^2-8x+9\) at (3,-6)
   6. to the curve \(y=(x^2-1)(x-2)\) at the points where the curve cuts the x-axis
3. Find the equation of the tangent and normal line to the graph of \(f(x)=\sqrt{x^2+3}\) at the point (−1,2).

---

---

---

Examples : Activity 2

1. Find the tangent to a curve \(y(x-2)(x-3)-x+7=0\) at a point where it meets x-axis
   

   Solution 👉 Click Here

   
   Given equation of curve is
   \(y(x-2)(x-3)-x+7=0\) (1)
   The curve meets x-axis at (x,0), therefore
   we put y=0
   \(y(x-2)(x-3)-x+7=0\)
   or \(0.(x-2)(x-3)-x+7=0\)
   or \(-x+7=0\)
   or \(x=7\)
   It shows that, the curve meets the x-axis at (7,0).
   Now, equation of tangent line at \((x_1,y_1)=(7,0)\) is
   \( y-y_1=m(x-x_1)\)
   or \( y-0=m(x-7)\)
   or \( y-0=\frac{dy}{dx}(x-7)\)(2)
   We know that given curve is
   \(y(x-2)(x-3)-x+7=0\)
   or \(y(x-2)(x-3)=x-7\)
   or \(y=\frac{x-7}{(x-2)(x-3)}\)
   or \(\frac{dy}{dx}=\frac{d}{dx} \frac{x-7}{x^2-5x+6}\)
   or \(\frac{dy}{dx}= \frac{\left [ \frac{d}{dx} (x-7) \right ] (x^2-5x+6)- \left [ \frac{d}{dx}(x^2-5x+6) \right ] (x-7)}{(x^2-5x+6)^2} \)
   or \(\frac{dy}{dx}= \frac{1. (x^2-5x+6)- (2x-5)(x-7)}{(x^2-5x+6)^2} \)
   At x=7, we get
   \(\frac{dy}{dx}= \frac{1. (7^2-5.7+6)}{(7^2-5.7+6)^2} \)
   or \(\frac{dy}{dx}= \frac{1}{20} \)
   Therefore, using (2) the required equation is
   \( y-0=\frac{dy}{dx}(x-7)\)
   or\( y-0=\frac{1}{20}(x-7)\)
   or x-20y=7
   This completes the solution
2. Find the horizental and vertical tangent to a curve \(x^2+y^2-xy-27=0\)
   

   Solution 👉 Click Here

   
   Given equation of curve is
   \(x^2+y^2-xy-27=0\) (1)
   Thus \(\frac{dy}{dx}=\frac{y-2x}{2y-x} \)
   Therefore, for horizental tangents, we set \(\frac{dy}{dx}=0\)
   Hence, we get
   y=2x(2)
   Solving (1) and (2), we get
   (3,6) and (-3,-6)
   Thus, taking (3,6) and (-3,-6) and \(\frac{dy}{dx}=0\), the horizental tangents are
   \( y=\pm 6\)
   Again
   For vertical tangents, we set \(\frac{dx}{dy}=0\)
   Hence, we get
   x=2y(3)
   Solving (1) and (3), we get
   (6,3) and (-6,-3)
   Thus, taking (6,3) and (-6,-3) and \(\frac{dx}{dy}=0\), the vertical tangents are
   \( x=\pm 6\)
   This completes the solution
3. Find the the points on a curve \(y=x^2+3x+4\), the tangents at which pass through the origin
   

   Solution 👉 Click Here

   
   Given equation of curve is
   \(y=x^2+3x+4\)
   Let the required point is (h,k)
   Then,
   \(k=h^2+3h+4\) (1)
   Also
   \(\frac{dy}{dx}=2x+3=2h+3\) at (h,k)
   Now,
   The equation of tangent at (h,k), which also pass through origin is
   \(k=\frac{dy}{dx} h \)
   or \(k=2h^2+3h \)(2)
   Now, Solving, (1) and (2), the required points are
   (2,14) and (-2,2)
   This completes the solutiuon
4. Find the point at which the tangent to the curve \(y=\sqrt{4x-3}-1\) has its slope \(\frac{2}{3}\)
5. Find the equation of all lines having slope 2 and being tangent to the curve \(y+\frac{2}{x-3}=0\). Also find the equations of normal at the points of contact.
6. Find the points on the curve \(y=x^3-11x+5\) at which equation of tangent is \(y=x-11\)
7. Find the equation of tangent and normal to the curves \(x=a \sin^3 \theta\) and \(y=a \cos^3 \theta\) at \(\theta =\frac{\pi}{4}\)
8. If a tangent line to \(y=f(x)\) at (4,3) passes through the point (0,2), find f(4) and f'(4).

---

---

---

Examples : Activity 3

Find the tangent and normal to a curve

1. \(x^2+2y^2=9\) at (1,2)
2. \(x^2+y^2=2\) at (1,1)
3. \(xy+2x-5y=2\) at (3,2)
4. \((y-x)^2=2x+4\) at (6,2)
5. \(x+\sqrt{xy}=6\) at (4,1)