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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 = âˆž.

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^{ }(x_{1}^{ },^{ }y_{1}) 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 = âˆž or dx/dy = 0.

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

**5. **For equation of tangent at (x_{1}, y_{1}), substitute xx_{1} for x^{2}, yy_{1} for y^{2}, for x, for y and for xy and keep the constant as such. This method is applicable only for second degree curves, i.e., ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

**6. **Method to find normal at (x_{1}, y_{1}) of second degree conics ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 The equation of normal at (x_{1}, y_{1}) is

**Ex.1 Find the equation of tangent to the ellipse 3x ^{2}+y^{2}+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) Ã— (slope of normal) = â€“1

The given equation 3x^{2} + y^{2} + x + 2y = 0 ....(iii)

Let (x_{1}, y_{1}) 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,

3x_{1}^{2} + 36x_{1}^{2} + x_{1} + 12x_{1} = 0

i.e., 13x_{1 }(3x^{1} + 1) = 0 â‡’ x_{1} = 0, â€“1/3

Using (iv), x_{1 }= 0 â‡’ y_{1} = 0 and x_{1} = â€“1/3 â‡’ y_{1} = â€“2

Hence, the points where tangent has slope â€“1/2 are P(0, 0) and Q(â€“1/3, â€“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 xâ€“axis y = 0, x + 0 = x^{0} â‡’ x = 1

Point is A(1, 0)

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

â‡’ log(x + y) = y log x

â‡’

slope of normal = 1

Equation of normal is,

â‡’ y = x â€“ 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 = 2x^{2} + 2

or 2x^{2} + 2x â€“ x â€“ 1 = 0 or (2x â€“ 1) (x + 1)

(If 2x^{2 }+x + 1 = 0 then x is imaginary)

From (1), for x = 1/2,

**Ex.4 Show that the curve x = 1 - 3t ^{2}, y = t - 3t^{3} 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 Ï† to OX. Prove that 3t = tanÏ† + secÏ†. 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 - 3t^{2} ...(1)

& y = t - 3t^{3} ...(2)

From (1) and (2), y = tx or â‡’ x^{3} = x^{2} - 3y^{2 }

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

For x > 1 â‡’ 1 â€“ 3t^{2} > 1 â‡’ â€“3t^{2} > 0 Impossible

From (1) and (2),

...given (3)

Adding (3) and (4) we get, tan Ã¸ + secÃ¸ = 3t

P(â€“2, 2)

1 â€“ 3t^{2} = â€“2 and 2 = t â€“ 3t^{3} then we get t = â€“1

Equation of tangent at (â€“2, 2) is

Therefore the tangent at t = â€“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 = x ^{3} 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 = x^{3} we get x^{3} - 12x + 16 = 0

this cubic must give all points of intersection of line and curve y = x^{3}

i.e., point P and Q.

But, since line is tangent at P so x = 2 will be a repeated root of equation x^{3} -12x + 16 = 0 and another root will be x = h.

Using theory of equations sum of roots â‡’ 2 + 2 + h = 0 â‡’ 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 Ï† with the axis of x, show that its equation is y cosÏ† - x sinÏ† = a cos 2Ï†.**

**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 + tan2Ã˜) = a^{2/3} Ãž y^{2/3} = a^{2/3} cos2Ã˜

y = a cos^{3} Ã˜ and x = a sin^{3} Ã˜

Therefore equation of normal is y â€“ a cos^{3} Ã˜ = tan Ã˜ (x â€“ a sin^{3} Ã˜)

y cos Ã˜â€“ a cos^{4} Ã˜ = x sin Ã˜ â€“ a sin^{4} Ã˜

y cosÃ˜â€“ x sin Ã˜ = a (cos^{4} Ã˜â€“ sin^{4} Ã˜)

= a (cos^{2} Ã˜ + sin^{2} Ã˜) (cos^{2} Ã˜- sin^{2} Ã˜) = a . 1 . cos 2Ã˜

Hence y cos Ã˜â€“ x sin Ã˜ = a cos 2Ã˜

**Ex.7 (a) Find y' if x ^{3} + y^{3} = 6xy.**

**(b) Find the tangent to the folium of Descartes x ^{3} + y^{3} = 6xy at the point (3, 3).**

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

**Sol. (a)** Differentiating both sides x^{3} + y^{3} = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y^{3} term and the Product Rule on the 6xy term, we get

3x^{2 }+ 3y^{2}y' = 6y + 6xy' or x^{2} + y^{2}y' = 2y + 2xy'

We now solve for y' :y^{2}y' â€“ 2xy' = 2y â€“ x^{2}

(y^{2 }â€“ 2x)y' = 2y â€“ x^{2},

(b) when x = y = 3

So, tangent to the folium of Descarte**s ** is y â€“ 3 = â€“1(x â€“ 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 â€“ x^{2} = 0. Substituting y = 1/2x^{2} , n the equation of the curve, we get

which simplifies to x^{6} = 16x^{3}. so either x = 0 or x^{3 }= 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(x_{1}, y_{1}) be the point of contact of the tangent.

Here, x^{a}y^{b} = k^{a + b }

âˆ´ a log x + b log y = (a + b) log k.

Solving with y = 0,

Let P divide AB in the ratio Î» : 1. Then

P divides AB in the constant ratio b : a.

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