A line can be called a tangent to the differentiable curve, y = f(x) at a point P if it makes an angle Θ at the x-axis. The derivative dy / dx = tan Θ is the slope of the tangent to the curve at a point.
Equation of normal at (x1, y1) is
where the slope is calculated at the point (x, y).
Few Important Points
- The slope of the tangent of the curve is given by the derivative of the curve equation with respect to x.
- The slope of the normal is given by
- The tangent equation is given by (y – b) =
- If x2 + y2 + 2gx + 2fy + c = 0 is a circle and a point is taken outside the circle such that a line drawn from the point to proceed through the boundary of a circle by coming in contact with the tangency point is given by,
Subtangent And Subnormal Definition
Let y = f (x) be the curve that is differentiable at a point P. Let the tangent and normal at P(x, y) to the curve meet at the x-axis at points T and N. M is the projection of P on the x-axis. In the figure below,
- PT is the length of the tangent
- PN is the length of the normal
- TM is the length of subtangent
- MN is the length of the subnormal
Subtangent And Subnormal Formulae
The formula for subtangent and subnormal is given below.
- The length of the tangent PT =
- The length of the subtangent PM =
- The length of the normal PN =
- The length of the subnormal MN =
Subtangent And Subnormal Solved Examples
Example 1: What is the equation of the tangent to the curve y = be[−x/a] at the point where it crosses the y-axis?
y = be[−x/a] meets the y-axis at (0, b).
Again, y = be[−x/a] (−1 / a)
At (0, b), dy / dx = be0 (−1 / a) = −b / a
Therefore, the required tangent is y − b = −b / a (x−0) or x / a + y / b = 1
Example 2: If x + 4y = 14 is normal to the curve y2 = ax3 − β at (2, 3), then find the value of α+β.
y2 = ax3 − β or dy / dx = 3αx2 / 2y
Therefore, slope of the normal at (2, 3) is (−dx / dy) (2,3) = −2 × 3 / 3α (2)2 = −12α = −14 Or α = 2 Also, (2, 3) lies on the curve.
Therefore, 9 = 8α − β or β = 16 − 9 = 7
α+β = 9
Example 3: If the normal to the curve is 2x2 + y2 = 12 at the point (2, 2) then at which of the following point does the curve cuts the normal again?
A) (−22 / 9, −29)
B) (22 / 9, 2 / 9)
D) None of these
2x2 + y2 = 12 or dy / dx = −2x / y.
The slope of normal at point A (2, 2) is 1 / 2.
Also, point B (−22 / 9,−2 / 9) lies on the curve and slope of AB is[2−(−2/9)] / [2−(−22/9)] = 1 / 2
Hence, the normal meets the curve again at point (−22 / 9,−2 / 9).
Example 4: The lines tangent to the curves y3 − x2 y + 5y − 2x = 0 and x4 − x3y2 + 5x + 2y = 0 at the origin intersect at an angle θ equal to ______.
Differentiating y3 − x2 y + 5y − 2x = 0 w.r.t. x,
we get 3 y2 y′− 2xy − x2y′ + 5y′ − 2 = 0
Or y′ = [2xy + 2] / [3y2 − x2 +5]
or y′(0,0) = 2 / 5
Differentiating x4 − x3y2 + 5x + 2y = 0 w.r.t. x,
we get 4x2 − 3x2y2 − 2x3yy′ + 5 + 2y′= 0
Or y′= [3x2y2 −4x3 − 5] / [2 − 2x3y]
or y′(0,0) = −5 / 2
Thus, both the curves intersect at a right angle.
Example 5: If the length of subnormal is equal to the length of subtangent at a point (3, 4) on the curve y = f (x) and the tangent at (3, 4) to y = f (x) meets the coordinate axes at A and B, then calculate the maximum area of the triangle OAB, where O is origin.
Length of subnormal = length of the subtangent
or dy/ d x = ±1
If dy / dx = 1, equation of the tangent is y − 4 = x − 3 or y − x = 1
The area of ΔOAB = [1 / 2] × 1 × 1 = 1 / 2
If dy / dx = −1, the equation of the tangent is y − 4 = −x + 3 Or y + x = 7,
Area = [1 / 2] × 7 × 7 = 49 / 2
Example 6: For the curve
Example 7: If ST and SN are the lengths of the subtangent and the subnormal at the point
Length of sub-tangent ST =
Length of the sub-normal SN =
Hence ST = SN.
Example 8: For the curve
Example 9: If the curve
Clearly the point of intersection of curves is (0, 1).
Now, the slope of the tangent of the first curve
Slope of tangent of second curve