Tangent is a straight line which passes through the curve or we can say the line touching the edge of a Circle or curve at a single Point, and never crosses it, there can be infinite number of Tangent to a circle or a curve.
Normal is the perpendicular Straight Line on the tangent at the same point where the tangent touches the curve. The derivative of the point where the tangent passes through the curve is its Slope. The equation of tangent is as under: y - y0 = k (x - x 0) here 'k' is gradient or Slope.
Tangents and Normals are both interlinked however it is not necessary for a normal to make a tangent but to make a normal a tangent is required. At any point suppose (r, t) we can find slope of the tangent with the use of dt / dr. Normal is another term for the perpendicular. When two lines meet each other at Right Angle or 90othen they are normal to each other. Let us say we have two lines MN & YZ, if they form an angle of 900 with each other, then we can say that line MN is normal to line YZ, and, also line YZ is normal to line MN.
With consideration in the formula which is mentioned above, we can take two gradients suppose k1 and k2 they both are perpendicular then k1k2 = -1. To find the equation of normal two gradients must be available. We can also calculate differential equation of the tangent and normal of a curve. This is all about tangents and normals.
Slope of TangentBack to Top
To understand about the Slope of Tangent, we should first be familiar about the tangents and slopes. Tangent to a curve at a Point is a straight line that touches the curve only at one point. The Slope of a line can be defined as, “the Ratio of change in Y- axis (vertical axis) to that of change in X - axis (horizontal axis)”. In mathematical terms, slope is denoted by ‘m’ and can be stated as m = (change in y) / (change in x).
Now coming back to Slope of tangent line: Let the curve is denoted by y = f(x), then the slope of the tangent line at point (x1, y1) can be given as the derivative of ‘y’ with respect to ‘x’ i.e. (dy/dx)x1,y1. With the help of point - slope method, the tangent line equation at point x1, y1 can be represented as (y - y1) = m*(x - x1), where ‘m’ is the slope of the tangent line. More explicitly denoting ‘m’ can be represented as m = (y - y1)/(x - x1). By seeing the mathematical formula for the slope of the curve, we can say that if the change in ‘x’ is not observable then slope can’t be defined at that point.
Mathematicians have also found out that there are few Functions and points on the graphs for which the slope doesn’t exist. The main reason behind this is the non-differentiability of the function at that point. But there are still few more reasons which are described below:
Let the function f(x) = x1/3, now we are trying to find the slope of the tangent at origin. Differentiating the given function gives (1/3)* x (-2/3) at x=0. Now as ‘x’ approaches to zero this value goes to infinity. Hence the slope of the tangent line is not defined and it can be said that tangent is a vertical line.
Now suppose f(x) = |x|. The slope at origin gives two slopes at a particular point. Moving from right of the origin we find the slope is 1 while moving from left of the origin gives us -1 as slope. Hence there is no unique tangent at this point and slope is not defined.
Differentiability suggests continuity, similarly discontinuity suggests non-differentiability. Hence we can derive that the slope of the tangent at a point, where the function is discontinuous can’t exist.
Tangents and Normals to a CurveBack to Top
A Tangent is basically a straight line to a plane curve. This line is at a fixed point on the curve. These tangent lines just touch the plane of curve very slightly. This line coincides at curve with the point. We can describe tangent with an equation also that is y = f (x). This line equation is basically valid for the point x = c. This point 'c' is any point at the curve if and only if line touches the curve. The point where line of tangent and curve meet is called as Point of Tangency. This point of tangent is extensively used in mathematics.
We can calculate the point of tangent in form of Slope by using two points 'x' and 'y' by dy / dx.
Finding normals is also a technique in Geometry mathematics when we want to find the movement of the object. Like radius of the Circle is always normal to the circle. Sometimes normal is used in place of perpendicular.
A normal is also a Straight Line to the circle. We use this normal line to find Intersection points on the circle. We can also calculate Slope of normal line. If there are two slopes of a curve say m1 and m2 then equation of normal line will be m1 * m2 = -1. These two slopes m1 and m2 are perpendicular to each other.
Tangents and Normals to a CircleBack to Top
Tangent to a circle can be defined as a line which touches the circumference of circle at single particular Point. There can be infinite tangents of circle.
If tangents are extended then they intersect with normal and meet Radius of Circle on same point making right angles. It means Tangents and Normals are perpendicular to each other that’s why they make 90 degree angle. Their multiplication results in -1.
Now we will Understand Tangents and Normals to a circle with help of a solved example where we have to find the equation of Tangent to the circle (x – 3)2 + (y – 2)2 = 25 at point (8, 4).
Solution: First step is to find the center with help of circle equation which are (3, 2).
Now we will find the gradient of normal to circle:
m= 4 - 2/ 8 – 3 = 2 / 5,
We know that gradients of two Perpendicular Lines if multiplied result in -1. So gradient of tangent will be -5 / 2. In given equation the tangent is passing through the points (8, 4). Equation of line through a point (x, y) will be (Y – y) = m (X – x). Now substitute these values:
Y - 4= -5 / 3 (x – 8),
Y= -5x/3 + 40/3 + 4,
Y=-5x/3 + (40 + 12) / 3,
Y=-5x/3 + 52/3,
Y= (52 – 5x) / 3.
Types of tangents:
1. Direct Common tangent: It is for two circles which meet on Center line and break it in the Ratio of radii externally.
2. Transverse Common tangent: These types of tangents meet at center line and breaks it in ratio of the radii internally.
Tangents and Normals DifferentiationBack to Top
Let’s say that tangent touches the curve at some point (X, Y); we can find the equation of tangent at this point. Slope of tangent can be calculated by taking differential that is by using dY / dX. Equation for normal to a curve can be written by calculating its Slope from condition for two lines with slopes “m” and “M” being perpendicular to each other as:
m * M = − 1,
For example, let us consider a curve: y = x3 – 2x2 + 5. Tangent touches the curve at a point (2, 5). Normal to the curve is perpendicular to tangent at same point. To find the derivative we use dy / dx:
Calculating m = d (x3 – 2 x2 + 5) / dx (at x = 2) = 3 x2 – 4x + 0 = 3 (2)2 – 4 * 2 + 0 = 4. Here m = 4 is the Slope of Tangent. To calculate slope of the normal i.e. “M” we can use formula: m * M = - 1
Or m = - 1 / 4.
So, equation of the normal can be written as: y – y1 = M (x – x1),
Where, y1 = 5, x1 = 2 and M = (-1 / 4).
So we can write above equation as: x + 4y – 22 = 0.
Tangents to a Curve From a Point Not on the CurveBack to Top
In case of ellipse the tangent’s Slope and intercept are bound to follow a certain tangential condition, according to which we have the following equation:
A2 w2 + B2 = c2 …....equation 1
In this equation 'A' represents the length of half of the major axis and 'B' represents the half the length of minor axis. “w” denotes the Slope of Tangent and c' 'denotes the intercept in tangent equation.
Example: Suppose there is an ellipse that is represented by an equation: h2 / 25 + k2 / 4 = 1, which has a tangent drawn from some arbitrary point (-3, 8/5). Find the equation of tangent?
Solution: Let equation of tangent be represented as k = v h + c,
Slope of tangent 'v' and its intercept would be following the condition for tangency, so we can write from equation 1 as:
A2 v2 + B2 = c2 ….....equation 2
Also the point (4, 5) lies on the tangent and thus it satisfies the equation:
8/5 = v (-3) + c,
Or c =8/5 + 3v,
Substituting the value of c, A and B in equation 2, we get:
25v2 + 4 = 64/25 + 9v2 + 48/5v,
'v' can be solved by using Quadratic Equation formula and then we can get 'c' also. This gives us the complete equation of tangent as:
k = 3/10 h + 5/2.
Equations of Tangents and Normals to the Conic SectionsBack to Top
To see how to solve such problems, let us consider some examples of conics:
Example 1: Find the equation of Tangent and normal at Point (2, 4) to the ellipse whose equation is given as follows:
X2 + 5Y2 = 40,
Solution: Slope of tangent can be found by differentiating the equation of ellipse as:
X2 + 5Y2 = 40,
On differentiating the equation we get:
2 X + 10 Y (dY / dX) = 0,
Or dY / dX = -2 X / 10 Y …....equation 1
Point at which the tangent touches the ellipse is given as: (2, 4).
Substituting the values of X and Y in equation 1, as X = 2 and Y = 4 we get the Slope of tangent as:
dY / dX = -2 X / 10 Y = -2 * 2 / 10 * 4 = -1 / 10,
Slope of normal can be calculated from the formula:
Slope (Tangent) * Slope (Normal) = - 1,
We get slope (normal) = -1 / (-1 / 10) = 10,
So, equations of tangent and normal to the ellipse can be written as follows:
(Y – 4) = -1 / 10 (X – 2),
Or Y = -X / 10 + 21 / 5,
(Y – 4) = 10 (X – 2),
Or Y = 10 X – 16,
In this way we will be differentiating the equations of other conics also and will be finding the Equations of Tangents and Normals at different points using point – slope form of line.
Equations of Tangents and NormalsBack to Top
We know that equation of line that passes through the point (x1, y1) with Slope 'm' is given as: (y – y1) = m (x – x1) hence the equation of tangent at point y = f (x) will be given by substituting y = f (x1) and m = f’ (x1). Thus (y – f (x1) = f’ (x1) (x – x1). Here f’ (x1) represents the first Derivative of Tangent at point x1.
Normal is defined as the line which is perpendicular to the tangent of the curve. For equation y = f (x), normal to the point y = f (x1) will be given by following formula in which the slope m is f’ (x1). In equation of normal line, we use 1 / f’ (x1) in place of f’ (x1) in the equation of tangent.
Thus the formula for the normal equation will be: (y – f (x1) = 1 / f’ (x1) (x – x1).
For example to find the equations of Tangents and Normals to the curve x = y2 + 1 at y = 3, we differentiate the given equation with respect to 'x' and get:
f’ (x) = 2y,
Thus at y = 3, f (x) = 10 and f’ (x) = 6. Equation of tangent will be written as:
(x – 10) = 6 (y – 3),
X – 6y – 7 = 0. And equation of the normal line will be
(x – 10) = (1 / 6) (y – 3),
6 x – y – 57 = 0.
Properties of Tangents and NormalsBack to Top
Derivatives of tangents and normals can be found to get the values of their respective slopes.
There is a relation that we define between the slopes of tangent and normal. Suppose a tangent touches the curve at some arbitrary point (a, b). Equation of this tangent can be written using the point – slope form of equation. Let slope of the tangent be “V”, equation can be written as:
(Y – b) = V (X – a)........... equation 1
As we have seen that normal and tangent are perpendicular to each other (Can be assumed as radius in the case of Circle), we have the relation:
VTANGENT * VNORMAL = - 1,
Or VNORMAL = - 1 / VTANGENT
So, only difference in the equation of normal as compared to that of tangent at a particular point is the slope. Slope of normal can be given by above formula and equation of normal can be written as follows:
(Y – b) = - 1 / VTANGENT (X – a),
For all curves the relation between tangent and normal of orthogonality will remain same.