If a line and a curve intersect at a point only then the line said to be tangent to the curve at that point. And the point is called the point of contact.
Tangents at (x, y) to y = f(x)
Lt y=f(x) be a given curve and P(x, y) and Q(x + δx, y + δy) be two neighbouring points on it
Equation of the line PQ is
Y – y = y + δy – y/x + δx – x(X – x)
Or Y – y = δy/δx(X – x) …(1)
Equation of line PQ
Equation of the line PQ
The line (1) will be a tagent to the give curve at P if Q → P which in turn means that δx → 0 and we know that
Lim δy/δx = dy/dx
δx → 0
Therefore the equation of the tangent is
Y – y = (dy/dx) (X – x)
Geometrical Intrepetation - slope and tangent
Geometrical meaning of dy/dx
From equation (1) we observe that dy/dx represents the slope of the tangent to the given curve y = f(x) at any point (x, y)
... dy/dx = tan Ψ
Where Ψ is the angle which the tangent to the curve makes with +ive direction of x-axis. In case we are to find the tangent at any point (x1, y1) then (dy/dx)(x1, y1) i.e. the value of dy/dx at (x1, y1) will represent the slope of the tangent and hence its equation in this case will be
y – y1 = (dy/dx)(x1, y1) (x – x1)
Algorithm to draw tangent of a circle
Algorithm to obtain the equations of tangents draw from a given point to a given circle.
Step i. Obtain the point, say (x1, y1)
Step ii. Write a line passing through (x1, y1) having slow m i.e., y – y1 = m (x – x1)
Step iii. Equate the length of the perpendicular from the centre of the circle to the line in step ll to the radius of the circle
Step iv. Obtain the value of m from the equation in step iii
Step v. Substitute m in the equation in step ll
Normal to the tangent:
The normal to a curve at a given point is a straight line passing through the given point, perpendicular to the tangent at this point. From the definition of a normal it is clear that the slope of the normal m′ and that of the tangent m are connected by the equation m′ = –`1/m .`
i.e., m′ = –1/f ′(x1)
= − 1/ dx/dy (x1, y1)
Hence the equation of a normal to a curve y = f(x) at a point P(x1,y1) is of the form y – y1= –1/f ′(x1)( x – x1).
The equation of the normal at (x1,y1) is
(i) x = x1 if the tangent is horizontal
(ii) y = y1 if the tangent is vertical and
(iii) y – y1 = –1 / m (x – x1) otherwise.
Example problems:
Example 1 : Find the equations of the tangents through (7, 1) to the circle x2 + y2 = 25
Solution : The equation of any line through (7, 1) is
y – 1 = m (x – 7)
mx – y – 7m + 1 = 0 …(i)
The coordinates of the centre and radius of the given circle are (0, 0) and 5 respectively
The line (i) will touch the given circle if,
Length of the perpendicular from the centre = radius
| (1 - 7m)/√(m2 + (-1)2 ) | = 5
│(1 – 7m)/√(m2 + 1)│ = 5
(1 – 7m)2/(m2 + 1) = 25
24m2 – 14m – 24 = 0
12m2 – 7m – 12 = 0
(4m + 3) (3m – 4) = 0
m = –3/4, 4/3
Substituting the values of m in (i), we obtain
3/4x – y + 21/4 + 1 = 0 and 4/3x – y – 28/3 + 1 = 0
3x + 4y – 25 = 0 and 4x – 3y – 25 = 0 which are the required equations.
Example 2: Find the equation of the tangent and normal to the curve y = x3 at the point (1,1).
Solution:
We have y = x3 ; slope y′= 3x2.
At the point (1,1), x = 1 and m = 3(1)2 = 3.
Therefore equation of the tangent is y − y1 = m(x − x1)
y – 1 = 3(x – 1) or y = 3x – 2
The equation of the normal is y − y1 = −`1/m ` (x − x1)
y – 1 =–`1/3` (x – 1) or y = –`1/3` x +4/3 .
normal tangent line graph
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