Tangent and Normal JEE Notes | EduRev

A. Tangent & Normal 

Definition : The tangent line to the graph of f at the point P(a, f(a)) is

(1) the line on P with slope f'(a) if f'(a) exists ;

(2) the line x = a if  = ∞.

In neither (1) nor (2) holds, then the graph of does not have a tangent line at the point P(a, f(a)).

In case f'(a) exists, then y - f(a) = f'(a) (x - a)

is an equation of the tangent line to the graph of f at the point P(a, f(a)).

The normal line N to the graph of a function f at the point P(a, f(a)) is defined to be the line through P perpendicular to the tangent line.

It follows that if f'(a)  0 the slope of N is -1/f'(a) and

 is an equation of N.

If f'(a) = 0, then N is the vertical line x = a; and if the tangent line is vertical, then N is the horizontal line y = f(a).

Note :

1. The point P (x₁ , y₁) will satisfy the equation of the curve & the equation of tangent & normal line.

2. If the tangent at any point P on the curve is parallel to the axis of x  then  dy/dx = 0 at the point P.

3. If the tangent at any point on the curve is parallel to the axis of y, then dy/dx = ∞ or dx/dy = 0.

4. If the tangent at any point on the curve is equally inclined to both the axes there dy/dx = ± 1.

5. For equation of tangent at (x₁, y₁), substitute xx₁ for x², yy₁ for y²,  for x,  for y and  for xy and keep the constant as such. This method is applicable only for second degree curves, i.e., ax² + 2hxy + by² + 2gx + 2fy + c = 0

6. Method to find normal at (x₁, y₁) of second degree conics ax² + 2hxy + by² + 2gx + 2fy + c = 0 The equation of normal at (x₁, y₁) is

 Ex.1 Find the equation of tangent to the ellipse 3x²+y²+x+2y = 0 which are perpendicular to the line 4x - 2y = 1.

 Sol. Since, tangent is the perpendicular to the line 4x - 2y = 1,

(slope of tangent) × (slope of normal) = –1

The given equation 3x² + y² + x + 2y = 0 ....(iii)

Let (x₁, y₁) be the point of contact of the tangent and the curve

From (i) and (iii), we get 

Substituting this in (ii) [since the points lies on the curve] we get,

3x₁² + 36x₁² + x₁ + 12x₁ = 0

i.e., 13x₁ (3x¹ + 1) = 0 ⇒ x₁ = 0, –1/3

Using (iv), x₁ = 0 ⇒ y₁ = 0 and x₁ = –1/3 ⇒ y₁ = –2

Hence, the points where tangent has slope –1/2 are P(0, 0) and Q(–1/3, –2).

Equation of tangents at P, Q are y  = -1/2x i.e. x+2y = 0

 i.e., 3x + 6y + 13 = 0 respectively..

Ex.2 Find the equation of normal to the curve x + y = x^y, where it cuts x-axis.

Sol. 

Given curve is x + y = x^y .....(i)

at x–axis y = 0,  x + 0 = x⁰ ⇒ x = 1

Point is A(1, 0)

Now to differentiation x + y = x^y taking log of both sides

⇒ log(x + y) = y log x

 ⇒

slope of normal = 1

Equation of normal is,  

⇒ y = x – 1

Ex.3 At what points on the curve  the tangents make equal angles with co-ordintae axes ?

Sol. 

Given curve is  ....(1)

Differentiating both sides w.r.t.x, then dy/dx = 2x² + 2

or 2x² + 2x – x – 1 = 0  or  (2x – 1) (x + 1) 

  (If 2x² +x + 1 = 0 then x is imaginary)

From (1), for x = 1/2, 

Ex.4 Show that the curve x = 1 - 3t², y = t - 3t³ is symmetrical about x-axis and has no real point for x > 1. If the tangent at the point t is inclined at an angle φ to OX. Prove that 3t = tanφ + secφ. If the tangent at P(-2, 2) meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Sol. Given curve is x = 1 - 3t²  ...(1)      

&  y = t - 3t³    ...(2)

From (1) and (2), y = tx   or      ⇒  x³ = x² - 3y² 

Since all powers of y are even, so curve is symmetrical about x-axis.

For x > 1 ⇒ 1 – 3t² > 1 ⇒ –3t² > 0 Impossible

From (1) and (2),

  ...given (3)

Adding (3) and (4) we get, tan ø + secø = 3t

P(–2, 2) 

1 – 3t² = –2 and 2 = t – 3t³ then we get t = –1

Equation of tangent at (–2, 2) is 

Therefore the tangent at t = –1 meets the curve again at 

Hence the tangents at P and Q are at right angles.

Ex.5 Tangent at P(2, 8) on the curve y = x³ meets the curve again at Q. Find coordinates of Q.

Sol. Equation of tangent at (2, 8) is y = 12x - 16

Solving this with y = x³ we get x³ - 12x + 16 = 0

this cubic must give all points of intersection of line and curve y = x³

i.e., point P and Q.

But, since line is tangent at P so x = 2 will be a repeated root of equation x³ -12x + 16 = 0 and another root will be x = h.

Using theory of equations sum of roots ⇒ 2 + 2 + h = 0   ⇒ h = -4

Hence coordinates of Q are (-4, -64)

Ex.6 If the normal to the curve x^2/3 + y^2/3 = a^2/3 makes an angle φ with the axis of x, show that its equation is y cosφ - x sinφ = a cos 2φ.

Sol. Given curve is x^2/3 + y^2/3 = a^2/3 ...(1)

Differentiating both sides w.r.t.x, we get

From (1) and (2),  y^2/3 (1 + tan2Ø) = a^2/3 Þ  y^2/3 = a^2/3 cos2Ø

 y = a cos³ Ø and x = a sin³ Ø

Therefore equation of normal is y – a cos³ Ø = tan Ø (x – a sin³ Ø)

y cos Ø– a cos⁴ Ø = x sin Ø – a sin⁴ Ø

y cosØ– x sin Ø = a (cos⁴ Ø– sin⁴ Ø) 

= a (cos² Ø + sin² Ø) (cos² Ø- sin² Ø) = a . 1 . cos 2Ø

Hence y cos Ø– x sin Ø = a  cos 2Ø

Ex.7 (a) Find y' if x³ + y³ = 6xy.

(b) Find the tangent to the folium of Descartes x³ + y³ = 6xy at the point (3, 3).

(c) At what points on the curve is the tangent line horizontal ?

Sol. (a) Differentiating both sides x³ + y³ = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y³ term and the Product Rule on the 6xy term, we get

3x² + 3y²y' = 6y + 6xy' or x² + y²y' = 2y + 2xy'

We now solve for y' :y²y' – 2xy' = 2y – x²

(y² – 2x)y' = 2y – x², 

(b) when x = y = 3

So, tangent to the folium of Descartes  is y – 3 = –1(x – 3)    or x + y = 6 

(c) The tangent line is horizontal if y' = 0. Using the expression for y' from part (a), we see that y' = 0 when 2y – x² = 0. Substituting y = 1/2x² , n the equation of the curve, we get  

which simplifies to x⁶ = 16x³. so either x = 0 or x³ = 16. If x = 16^1/3 = 2^4/3, then y = 1/2(2^8/3) = 2^5/3. Thus, the tangent is horizontal at (0, 0) and at (2^4/3, 2^5/3).

Ex.8 In the curve x^a y^b = k^{a + b}, (a b > 0) prove that the portion of the tangent intercepted between the coordinate axes is divided at its point of contact into segments which are in constant ratio.

Sol. Let P(x₁, y₁) be the point of contact of the tangent.

Here, x^ay^b = k^{a + b} 

∴ a log x + b log y = (a + b) log k.

Solving with y = 0, 

Let P divide AB in the ratio λ : 1. Then 

P divides AB in the constant ratio b : a.