Tangent and Normal JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : Tangent and Normal JEE Notes | EduRev

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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 Tangent and Normal JEE Notes | EduRev = ∞.

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) Tangent and Normal JEE Notes | EduRev 0 the slope of N is -1/f'(a) and

Tangent and Normal JEE Notes | EduRev 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, Tangent and Normal JEE Notes | EduRev for x, Tangent and Normal JEE Notes | EduRev for y and Tangent and Normal JEE Notes | EduRev 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 Tangent and Normal JEE Notes | EduRev

 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

Tangent and Normal JEE Notes | EduRev

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

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

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

From (i) and (iii), we get 

Tangent and Normal JEE Notes | EduRev

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

3x12 + 36x12 + x1 + 12x1 = 0

i.e., 13x(3x1 + 1) = 0 ⇒ x1 = 0, –1/3

Using (iv), x= 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

Tangent and Normal JEE Notes | EduRev i.e., 3x + 6y + 13 = 0 respectively..

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

Sol. 

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

at x–axis y = 0,  x + 0 = x0 ⇒ x = 1

Point is A(1, 0)

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

⇒ log(x + y) = y log x

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev ⇒

Tangent and Normal JEE Notes | EduRev

slope of normal = 1

Equation of normal is,   Tangent and Normal JEE Notes | EduRev

⇒ y = x – 1

Ex.3 At what points on the curve Tangent and Normal JEE Notes | EduRev the tangents make equal angles with co-ordintae axes ?

Sol. 

Given curve is Tangent and Normal JEE Notes | EduRev ....(1)

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

Tangent and Normal JEE Notes | EduRev

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

Tangent and Normal JEE Notes | EduRev  (If 2x+x + 1 = 0 then x is imaginary)

From (1), for x = 1/2,  Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

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   Tangent and Normal JEE Notes | EduRev   ⇒  x3 = x2 - 3y

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),

Tangent and Normal JEE Notes | EduRev  ...given (3)

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

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

Tangent and Normal JEE Notes | EduRev

Equation of tangent at (–2, 2) is  Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Therefore the tangent at t = –1 meets the curve again at  Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

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.

Tangent and Normal JEE Notes | EduRev

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 x2/3 + y2/3 = a2/3 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 x2/3 + y2/3 = a2/3 ...(1)

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

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev
Tangent and Normal JEE Notes | EduRev

From (1) and (2),  y2/3 (1 + tan2Ø) = a2/3 Þ  y2/3 = a2/3 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

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

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

(y– 2x)y' = 2y – x2Tangent and Normal JEE Notes | EduRev

(b) when x = y = 3

Tangent and Normal JEE Notes | EduRev

So, tangent to the folium of Descarte 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   Tangent and Normal JEE Notes | EduRev

which simplifies to x6 = 16x3. so either x = 0 or x= 16. If x = 161/3 = 24/3, then y = 1/2(28/3) = 25/3. Thus, the tangent is horizontal at (0, 0) and at (24/3, 25/3).

Ex.8 In the curve xa yb = ka + 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, xayb = ka + b 

∴ a log x + b log y = (a + b) log k.

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Solving with y = 0, 

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

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

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

Tangent and Normal JEE Notes | EduRev

P divides AB in the constant ratio b : a.

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