Subtangent and Subnormal

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).

A Few Important Points

• The slope of the tangent of the curve is given by the derivative of the curve equation with respect to x.
• \(\begin{array}{l}\text{The slope of the normal is given by}\ \frac{-1}{\frac{dy}{dx}} (x_1, y_1).\end{array} \)
• The tangent equation is given by
  \(\begin{array}{l}(y – b) = \frac{-dx}{dy} (x-a)\end{array} \)
• 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
  \(\begin{array}{l}y-(-f)=m(x-[-g]*\sqrt{1+m^2})\end{array} \)

Subtangent and Subnormal Definition

Let y = f (x) be the differentiable curve 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 the 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
  \(\begin{array}{l}PT=|y|\sqrt{1+[\frac{dx}{dy}]^2}\end{array} \)
• The length of the subtangent
  \(\begin{array}{l}TM=|ycot\ \theta |=|y*\frac{dx}{dy}|\end{array} \)
• The length of the normal
  \(\begin{array}{l}PN=|y|\sqrt{1+[\frac{dy}{dx}]^2}\end{array} \)
• The length of the subnormal
  \(\begin{array}{l}MN=|ycot\ \theta |=|y*\frac{dy}{dx}|\end{array} \)

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?

Solution:

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.

Solution:

y² = ax³ − β or dy / dx = 3αx² / 2y

Therefore, the 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

or

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 cut the normal again?

A) (−22 9, −29)

B) (22/9, 2/9)

C) (−2,−2)

D) None of the above

Solution:

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 the slope of AB is

Hence, the normal meets the curve again at the 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 ______.

Solution:

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 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 the origin.

Solution:

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 by² = (x + a)³ the square of subtangent is proportional to

Solution: 

Example 7: If ST and SN are the lengths of the subtangent and the subnormal at the point θ = π/2 on the curve x = a(θ + sin θ), y = a(1 – cos θ), a ≠ 1, then

Solution:

The length of the subtangent ST

The length of the subnormal SN

Hence, ST = SN

Example 8: For the curve xy = c², the subnormal at any point varies as

A) x²

B) x³

C) y²

D) y³

Solution:

xy = c² …..(i)

Because subnormal = y(dy/dx)

From (i), y = c²/x

Subnormal varies as y³.

Example 9: If the curve y = a^x and y = b^x intersect at angle α, then find tan α.

Solution:

Clearly, the point of intersection of curves is (0, 1).

Now, the slope of the tangent of the first curve

The slope of the tangent of the second curve

Length of Tangent, Normal, Sub-Tangent & Sub-normal – Video Lesson