# RD Sharma Class 12 Ex 16.2 Solutions Chapter 16 Tangents and Normals

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## RD Sharma Class 12 Ex 16.2 Solutions Chapter 16 Tangents and

### Question 1. Find the equation of the tangent to the curve √x + √y = a at the point (a2/4, a2/4).

Solution:

We have,

√x + √y = a

On differentiating both sides w.r.t. x, we get

dy/dx = -√y/√x

Given, (x1, y1) = (a2/4, a2/4),

Slope of tangent, m =

The equation of tangent is,

y – y1 = m (x – x1)

y – a2/4 = –1(x – a2/4)

y – a2/4 = –x + a2/4

x + y = a2/2

### Question 2. Find the equation of the normal to y = 2x3 − x2 + 3 at (1, 4).

Solution:

We have,

y = 2x3 − x2 + 3

On differentiating both sides w.r.t. x, we get

dy/dx = 6x– 2x

Slope of tangent =  = 6 (1)2 – 2 (1) = 4

Slope of normal = – 1/Slope of tangent = – 1/4

Given, (x1, y1) = (1, 4),

The equation of normal is,

y – y1 = m (x – x1)

y – 4 = -1/4 (x – 1)

4y – 16 = – x + 1

x + 4y = 17

### (i) y = x4 − bx3 + 13x2 − 10x + 5 at (0, 5)

Solution:

We have,

y = x4 − bx3 + 13x2 − 10x + 5

On differentiating both sides w.r.t. x, we get

dy/dx = 4x3 – 3bx2 + 26x – 10

Slope of tangent, m=  = -10

Given, (x1, y1) = (0, 5)

The equation of tangent is,

y – y1 = m (x – x1)

y – 5 = – 10 (x – 0)

y – 5 = -10x

y + 10x – 5 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 5 = 1/10 (x – 0)

10y – 50 = x

x – 10y + 50 = 0

### (ii) y = x4 − 6x3 + 13x2 − 10x + 5 at x = 1

Solution:

We have,

y = x4 − 6x3 + 13x2 − 10x + 5

When x = 1, we have y = 1 – 6 + 13 – 10 + 5 = 3

So, (x1, y1) = (1, 3)

Now, y = x4 − 6x3 + 13x2 − 10x + 5

On differentiating both sides w.r.t. x, we get

dy/dx = 4 x3 – 18 x2 + 26x – 10

Slope of tangent, m =  = 4 – 18 + 26 – 10 = 2

The equation of tangent is,

y – y1 = 2 (x – x1)

y – 3 = 2 (x – 1)

y – 3 = 2x – 2

2x – y + 1 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 3 = -1/2 (x – 1)

2y – 6 = – x + 1

x + 2y – 7 = 0

### (iii) y = x2 at (0, 0)

Solution:

We have,

y = x2

On differentiating both sides w.r.t. x, we get

dy/dx = 2x

Given, (x1, y1) = (0, 0)

Slope of tangent, m=  = 2 (0) = 0

The equation of tangent is,

y – y1 = m (x – x1)

y – 0 = 0 (x – 0)

y = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 0 = -1/0 (x – 0)

x = 0

### (iv) y = 2x2 − 3x − 1 at (1, −2)

Solution:

We have,

y = 2x− 3x − 1

On differentiating both sides w.r.t. x, we get

dy/dx = 4x – 3

Given, (x1, y1) = (1, -2)

Slope of tangent, m =  = 4 – 3 = 1

The equation of tangent is,

y – y1 = m (x – x1)

y + 2 = 1 (x – 1)

y + 2 = x – 1

x – y – 3 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y + 2 = -1 (x – 1)

y + 2 = -x + 1

x + y + 1 = 0

### (v)  at (2, -2)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

Given, (x1, y1) = (2, -2)

Slope of tangent, m=  =  = -2

The equation of tangent is,

y – y1 = m (x – x1)

y + 2 = -2 (x – 2)

y + 2 = -2x + 4

2x + y – 2 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y + 2 = 1/2 (x – 2)

2y + 4 = x – 2

x – 2y – 6 = 0

### (vi) y = x2 + 4x + 1 at x = 3

Solution:

We have,

y = x2 + 4x + 1

On differentiating both sides w.r.t. x, we get,

dy/dx = 2x + 4

When x = 3, y = 9 + 12 + 1 = 22

So, (x1, y1) = (3, 22)

Slope of tangent, m=  = 10

The equation of tangent is,

y – y1 = m (x – x1)

y – 22 = 10 (x – 3)

y – 22 = 10x – 30

10x – y – 8 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 22 = -1/10 (x – 3)

10y – 220 = – x + 3

x + 10y – 223 = 0

### (vii)  at (a cos θ, b sin θ)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

dy/dx = -x b2/y a2

Slope of tangent, m=

Given, (x1, y1) = (a cos θ, b sin θ)

The equation of tangent is,

y – y1 = m (x – x1)

y – b sin θ = -bcosθ/asinθ  (x – a cos θ)

ay sin θ – ab sin2 θ = -bx cos θ + ab cos2 θ

bx cos θ + ay sin θ = ab

On dividing by ab, we get

x/a cosθ + y/b sinθ = 1

The equation of normal is,

y – y1 = -1/m (x – x1)

y – b sin θ = asinθ/bcosθ  (x – a cos θ)

by cos θ – b2 sin θ cos θ = ax sin θ – a2 sin θ cos θ

ax sin θ – by cos θ = (a2 – b2) sin θ cos θ

On dividing by sin θ cos θ, we get

ax sec θ – by cosec θ = a2 – b2

### (viii)  at (a sec θ, b tan θ)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

dy/dx = x b2/y a2

Slope of tangent}, m=

Given, (x1, y1) = (a sec θ, b tan θ)

The equation of tangent is,

y – y1 = m (x – x1)

y – b tan θ =  (x – a sec θ)

ay sin θ – ab(sin2 θ/cos θ) = bx – (ab/cos θ)

ay sin θ cos θ – ab sin2 θ = bx cos θ – ab

bx cos θ – ay sin θ cos θ = ab (1 – sin2 θ)

bx cos θ – ay sin θ cos θ = ab cos2 θ

On dividing by ab cos2 θ, we get

x/a sec θ – y/b tan θ = 1

The equation of normal is,

y – y1 = -1/m (x – x1)

y – b tan θ = -a sin θ/b (x – a sec θ)

by – b2 tan θ = -ax sin θ + a2 tan θ

ax sin θ + by = (a2 + b2) tan θ

On dividing by tan θ, we get

ax cos θ + by cot θ = a2 + b2

### (ix) y2 = 4ax at (a/m2, 2a/m)

Solution:

We have,

y2 = 4ax

On differentiating both sides w.r.t. x, we get

dy/dx = 2a/y

Given, (x1, y1) = (a/m2, 2a/m)

Slope of tangent =  = m

The equation of tangent is,

y – y1 = m (x – x1)

my – 2a = m2 x – a

m2 x – my + a = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

m3 y – 2a m2 = – m2 x + a

m2 x + m3 y – 2a m2 – a = 0

### (x) c2 (x2 + y2) = x2y2 at (c/cos θ, c/sin θ)

Solution:

We have,

c2 (x2 + y2) = x2y2

On differentiating both sides w.r.t. x, we get

2x c2 + 2y c2(dy/dx) = x2 2y(dy/dx) + 2x y2

dy/dx(2y c2 – 2 x2 y) = 2x y2 – 2x c2

Slope of tangent, m=

= -cos3 θ/ sin3 θ

Given, (x1, y1) = (c/cos θ, c/sin θ)

The equation of tangent is,

y – y1 = m (x – x1)

sinθ (y sin θ – c) = -cos2 θ (x cos θ – c)

y sin3 θ – c sin2 θ = – x cos3 θ + c cos2 θ

x cos3 θ + y sin3 θ = c ( sin2 θ + cos2 θ)

x cos3 θ + y sin3 θ = c

The equation of normal is,

y – y1 = -1/m (x – x1)

sin3 θ – ycos3 θ = 2c[-cos (2θ)/sin(2θ)]

sin3 θ – y cos3 θ = -2c cot 2θ

sin3 θ – y cos3 θ + 2c cot 2θ = 0

### (xi) xy = c2 at (ct, c/t)

Solution:

We have,

xy = c2

On differentiating both sides w.r.t. x, we get

dy/dx = – y/x

Given, (x1, y1) = (ct, c/t)

Slope of tangent, m=

The equation of tangent is,

y – y1 = m (x – x1)

yt2 – ct = -x + ct

x + y t2 = 2ct

The equation of normal is,

y – y1 = -1/m (x – x1)

y – c/t – t2(x – ct)

yt – c = t3 x – c t4

x t3 – yt = c t4 – c

### (xii) at (x1, y1)

Solution:

We have,

On differentiating both sides w.r.t. x,

dy/dx = – x b2/y a2

Slope of tangent, m=

The equation of tangent is,

y – y1 = m (x – x1)

y – y1 = – x1 b2/y1 a2(x – x1)

y y1 a2 – y12 a2 = -x x1 b2 + x12 b2

x x1 b2 + y y1 a2 = x12 b2 + y12 a2  . . . . (1)

Given (x1, y1) lies on the curve, we get

x12 b2 + y12 a2 = a2 b2

Substituting this in (1), we get

x x1 b2 + y y1 a2 = a2 b2

On dividing this by ab2, we get

The equation of normal is,

y – y1 = m (x – x1)

y – y1 = y1 a2/x1 b2(x – x1)

y x1 b2 – x1 y1 b2 = x y1 a2 – xy1 a2

x y1 a2 – y x1 b2 = x1 y1 a2 – x1 y1 b2

x y1 a2 – y x1 b2 = x1 y1 (a2 – b2)

On dividing by x1 y1, we get

### (xiii) at (x0 , y0)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

dy/dx = x b2/y a2

Slope of tangent, m=

The equation of tangent is,

y – y1 = m (x – x1)

y – y0 = x0 b2/y0 a2(x – x0)

y y0 a2 – y02 a2 = x x0 b2 – x02 b2

x x0 b2 – y y0 a2 = x02 b2 – y0a2  . . . . (1)

x02 b2 – y02 a2 = ab

Substituting this in eq(1), we get,

x x0 b2 – y y0 a2 = ab2

Dividing this by ab2, we get

The equation of normal is,

y – y1 = m (x – x1)

y – y0 = y0 a2/x0 b2(x – x0)

y x0 b2 – xy0 b2 = -x y0 a2 + x0 y0 a2

x y0 a2 + y x0 b2 = x0 y0 a2 + x0 y0 b2

x y0 a2 + y x0 b2 = x0 y0 (a2 + b2)

Dividing by x0 y0, we get

### (xiv) at (1, 1)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

Slope of tangent, m= = -1

Given, (x1, y1) = (1, 1)

The equation of tangent is,

y – y1 = m (x – x1)

y – 1 = -1 (x – 1)

y – 1 = -x + 1

x + y – 2 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 1 = 1 (x – 1)

y – 1 = x – 1

y – x = 0

### (xv) x2 = 4y at (2, 1)

Solution:

We have,

x2 = 4y

On differentiating both sides w.r.t. x, we get

2x = 4dy/dx

dy/dx = x/2

Slope of tangent, m=  = 2/2 = 1

Given, (x1, y1) = (2, 1)

The equation of tangent is,

y – y1 = m (x – x1)

y – 1 = 1 (x – 2)

y – 1 = x – 2

x – y – 1 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 1 = – 1 (x – 2)

y – 1 = – x + 2

x + y – 3 = 0

### (xvi) y2 = 4x at (1, 2)

Solution:

We have,

y2 = 4x

On differentiating both sides w.r.t. x, we get

2y (dy/dx) = 4

dy/dx = 2/y

Slope of tangent, m= = 2/2 = 1

Given, (x1, y1) = (1, 2)

The equation of tangent is,

y – y1 = m (x – x1)

y – 2 = 1 (x – 1)

y – 2 = x – 1

x – y + 1 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 2 = -1 (x – 1)

y – 2 = -x + 1

x + y – 3 = 0

### (xvii) 4x2 + 9y2 = 36 at (3 cos θ, 2 sin θ)

Solution:

We have,

4x2 + 9y2 = 36

On differentiating both sides w.r.t. x, we get

8x + 18y dy/dx = 0

18y dy/dx = – 8x

dy/dx = -8x/18y = -4x/9y

Slope of tangent, m =

The equation of tangent is,

y – y1 = m (x – x1)

y – 2 sin θ = -2 cos θ/3 sin θ(x – 3 cos θ)

3y sin θ – 6 sin2 θ = -2x cos θ + 6 cos2 θ

2x cos θ + 3y sin θ = 6 (cos2 θ + sinθ)

2x cos θ + 3y sin θ = 6

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 2 sin θ = -3 sin θ/2 cos θ(x – 3 cos θ)

2y cos θ – 4 sin θ cos θ = 3x sin θ – 9 sin θ cos θ

3x sin θ – 2y cos θ – 5sin θ cos θ = 0

### (xviii) y2 = 4ax at (x1, y1)

Solution:

We have,

y2 = 4ax

On differentiating both sides w.r.t. x, we get

2y dy/dx = 4a

dy/dx = 2a/y

At (x1, y1), we have

Slope of tangent = = m

The equation of tangent is,

y – y1 = m (x – x1)

y y1 – y12 = 2ax – 2a x1

y y1 – 4a x1 = 2ax – 2a x1

y y1 = 2ax + 2a x1

y y1 = 2a (x + x1)

The equation of normal is,

y – y1 = -1/m (x – x1)

y – y1 = -y1/2a (x – x1)

### (xix) at (√2a, b)

Solution:

We have,

On differentiating both sides w.r.t. x, we get

dy/dx = x b2/y a2

Slope of tangent, m=

The equation of tangent is,

y – y1 = m (x – x1)

y – b = √2b/a(x – √2a)

ay – ab = √2 bx – 2ab

√2 bx – ay = ab

The equation of normal is,

y – y1 = -1/m (x – x1)

y – b = – a/√2b(x – √2a)

√2 by – √2 b2 = – ax + √2 a

ax + √2 by = √2 b2 + √2 a2

ax/√2 + by = a2 + b2

### Question 4. Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Solution:

We have,

x = θ + sin θ, y = 1 + cos θ

and

Slope of tangent, m =

= 1 – √2

Given, (x1, y1) = (π/4 + sin π/4, 1 + cos π/4) = (π/4 + 1/√2, 1 + 1/√2)

The equation of tangent is,

y – y1 = m (x – x1)

y – (1 + 1/√2) = (1 – √2) [x – (π/4 + 1/√2)]

y – 1 – 1/√2 = (1 – √2) (x – π/4 – 1/√2)

### (i) x = θ + sin θ, y = 1 + cos θ at θ = π/2

Solution:

We have,

x = θ + sin θ and y = 1 + cos θ

dx/dθ = 1 + cos θ and dy/dθ = -sinθ

Slope of tangent, m=

= -1/(1 + 0)

= -1

Given, (x1, y1) = (π/2 + sin π/2, 1 + cos π/2) = (π/2 + 1, 1)

The equation of tangent is,

y – y1 = m (x – x1)

y – 1 = -1 (x – π/2 – 1)

2y – 2 = – 2x + π + 2

x + 2y – π – 4 = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y – 1 = 1 (x – π/2 -1)

2y – 2 = 2x – π – 2

2x – 2y = π

### (ii)  at t = 1/2

Solution:

We have,

dx/dt =

dy/dt =

=

Slope of tangent, m=

Given, (x1, y1) =

The equation of tangent is,

y – y1 = m (x – x1)

80y – 16a = 65x – 26a

65x – 80y – 10a = 0

13x – 16y – 2a = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

65y – 13a = – 80x + 32a

80x + 65y – 45a = 0

16x + 13y – 9a = 0

### (iii) x = at2, y = 2at at t = 1

Solution:

We have,

x = at2, y = 2at

dx/dt = 2at and dy/dt = 2a

Slope of tangent, m=  = 1

Given, (x1 , y1) = (a, 2a)

The equation of tangent is,

y – y1 = m (x – x1)

y – 2a = 1 (x – a)

y – 2a = x – a

x – y + a = 0

Equation of normal:

y – y1 = -1/m (x – x1)

y – 2a = – 1 (x – a)

y – 2a = – x + a

x + y = 3a

### (iv) x = a sec t, y = b tan t at t

Solution:

We have,

x = a sec t, y = b tan t

dx/dt = a sect tant and dy/dt = b sec2t

Slope of tangent, m =

Given (x1, y1) = (a sec t, b tan t)

The equation of tangent is,

y – y1 = m (x – x1)

y – b tan t = (b/a) cosec t (x – a sec t)

ay sin t cos t – ab sin2 t = bx cos t – ab

bx cos t – ay sin t cos t – ab (1 – sin2 t) = 0

bx cos t – ay sin t cos t = ab cos2 t

On dividing by cos2 t, we get

bx sec t – ay tan t = ab

The equation of normal is,

y – y1 = -1/m (x – x1)

y – b tant = -a/b sint(x – asect)

ycost − bsint = − a/b​sint(xcost − a)

by cos t – b2 sin t = – ax sin t cos t + a2 sin t

ax sin t cos t + by cos t = (a2 + b2) sin t

On dividing both sides by sin t, we get

ax cos t + by cot t = a+ b2

### (v) x = a(θ + sin θ), y = a(1 − cos θ) at θ

Solution:

We have,

x = a(θ + sin θ), y = a(1 − cos θ)

dx/dθ = a(1 + cosθ) and dy/dθ = asinθ

= tan θ/2   . . . . (1)

Slope of tangent, m=

Given (x1, y1) = [a(θ + sin θ), a(1 − cos θ)]

The equation of tangent is,

y – y1 = m (x – x1)

y – a (1 – cos θ) = tan θ/2 [x – a (θ + sin θ)]

y − a(2sin2θ/2​) = xtanθ/2​ − aθtanθ/2 ​− atanθ/2​sinθ

y − 2asin2θ/2 ​ =(x − aθ)tan θ/2 − 2asin2θ/2​

y = (x – aθ) tan θ/2

The equation of normal is,

y – a (1 – cos θ) = -cot θ/2 [x – a (θ + sin θ)]

tanθ/2​[y − a(2sin2θ/2​)] = −x + aθ + asinθ

tanθ/2​[y − a{2(1 − cos2θ/2​)}] = −x + aθ + asinθ

tan θ/2 (y – 2a) + a (2sin θ/2 cosθ/2 = -x + aθ + a sinθ

tan θ/2 (y – 2a) + a sin θ = -x + aθ + a sin θ

tan θ/2 (y – 2a) = – x + aθ

tan θ/2 (y – 2a) + x – θ = 0

### (vi) x = 3 cos θ − cos3 θ, y = 3 sin θ − sin3 θ

Solution:

We have,

x = 3 cos θ − cos3 θ, y = 3 sin θ − sin3 θ

dx/dθ = -3sin θ + 3 cos2θ sin θ and dy/dθ = 3 cos θ – 3 sin2θ cos θ

=  cos3 θ/ -sin3 θ

= tan3 θ

So the equation of the tangent at θ is,

y – 3 sin θ + sinθ = -tanθ (x – 3 cos θ + cos3 θ)

4 (y cosθ – x sinθ) = 3 sin 4θ

So the equation of normal at θ is,

y – 3 sin θ + sin3 θ= (1/tan3 θ) (x – 3 cos θ + cos3 θ)

sinθ – x cos3 θ = 3sinθ – sin6 θ – 3cosθ + cosθ

### Question 6. Find the equation of the normal to the curve x2 + 2y2 − 4x − 6y + 8 = 0 at the point whose abscissa is 2?

Solution:

Given that abscissa = 2. i.e., x = 2

x2 + 2y2 − 4x − 6y + 8 = 0  . . . . (1)

On differentiating both sides w.r.t. x, we get

2x + 4y dy/dx – 4 – 6 dy/dx = 0

dy/dx(4y – 6) = 4 – 2x

When x = 2, we get

4 + 2y2 – 8 – 6y + 8 = 0

2y2 – 6y + 4 = 0

y2 – 3y + 2 = 0

y = 2 or y = 1

m (tangent) at x = 2 is 0

Normal is perpendicular to tangent so, m1m2 = –1

m (normal) at x = 2 is 1/0, which is undefined.

The equation of normal is given by y – y1 = m (normal) (x – x1)

x = 2

### Question 7. Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).

Solution:

We have,

ay2 = x3

On differentiating both sides w.r.t. x, we get

2aydy/dx = 3x2

dy/dx = 3x2/2ay

Slope of tangent =

Given (x1, y1) = (am2, am3)

The equation of normal is,

y – y1 = -1/m (x – x1)

y – a m3 = -2m/3 (x – am2)

3my – 3am4 = – 2x + 2am2

2x + 3my – am2 (2 + 3m2) = 0

### Question 8. The equation of the tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x − 5. Find the values of a and b.

Solution:

We have,

y2 = ax3 + b

On differentiating both sides w.r.t. x, we get

2y dy/dx = 3ax2

dy/dx = 3ax2/2y

Slope of tangent, m =

The equation of tangent is given by y – y1 = m (tangent) (x – x1)

Now compare the slope of a tangent with the given equation

2a = 4

a = 2

Now (2, 3) lies on the curve, these points must satisfy

32 = 2 (23) + b

b = – 7

### Question 9. Find the equation of the tangent line to the curve y = x2 + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

Solution:

We have,

y = x2 + 4x − 16

Let (a, b) be the point of intersection of both the curve and the tangent.

Since (a, b) lies on curve, we get

b = a2 + 4a − 16

Now, x2 + 4x − 16

dy/dx = 2x + 4

Slope of tangent =

Given that the tangent is parallel to the line we have,

Slope of tangent = Slope of the given line

=> 2a + 4 = 3

=> 2 a = -1

=> a = -1/2

From eq(1), we get

b = 1/4 – 2 – 16 = -71/4

Now, slope of tangent, m = 3

(a, b) = (-1/2, -71/4)

The equation of tangent is,

y – y1 = m (x – x1)

y + 71/4 = 3 (x + 1/2)

4y + 71 = 12x + 6

12x – 4y – 65 = 0

### Question 10. Find an equation of normal line to the curve y = x3 + 2x + 6 which is parallel to the line x+ 14y + 4 = 0.

Solution:

We have,

y = x3 + 2x + 6

Let (a, b) be a point on the curve where we need to find the normal.

Slope of the given line = -1/14

Since the point lies on the curve, we get

b = a3 + 2a + 6

Now, y = x+ 2x + 6

dy/dx = 3 x+ 2

Slope of the tangent, m =

Slope of the normal =

Given that, slope of the normal = slope of the given line, we have

3a2 + 2 = 14

3a2 = 12

a2 = 4

a = ±2

So, b = 18 or -6.

And slope of the normal = -1/14

When a = 2 and b = 18, we have

y – y= m (x – x1)

y – 18 = -1/14 (x – 2)

14y – 252 = -x + 2

x + 14y – 254 = 0

When a = -2 and b = -6, we have

y – y1 = m (x – x1)

y + 6 = -1/14 (x + 2)

14y + 84 = -x – 2

x + 14y + 86 = 0

### Question 11. Determine the equation(s) of tangent (s) line to the curve y = 4x3 − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Solution:

Let (a, b) be a point on the curve where we need to find the tangent(s).

Slope of the given line = -1/9

Since, tangent is perpendicular to the given line,

Slope of the tangent =  = 9

Hence, b = 4 a3 – 3 a + 5

Now, y = 4 x3 – 3x + 5

dy/dx = 12 x2 – 3

Slope of the tangent =

Given that, slope of the tangent = slope of the perpendicular line

12a2 – 3 = 9

12a2 = 12

a2 = 1

a = ±1

So, b = 6 or 4.

Thus, slope of tangent = 9.

When a = 1 and b = 6, we have

y – y1 = m (x – x1)

y – 6 = 9 (x – 1)

y – 6 = 9x – 9

9x – y – 3 = 0

When a = -1 and b = 4, we have

y – y1 = m (x – x1)

y – 4 = 9 (x + 1)

y – 4 = 9x + 9

9x – y + 13 = 0

### Question 12. Find the equation of a normal to the curve y = x loge x which is parallel to the line 2x − 2y + 3 = 0.

Solution:

Slope of the given line is 1.

Let (a, b) be the point where the tangent is drawn to the curve.

Hence, b = a loge a  . . . . (1)

Now, y = x loge x

dy/dx = x × 1/x + loge x(1) = 1 + loge x1

Slope of tangent = 1 + log a

Slope of normal =

Given that, slope of normal = slope of the given line.

=> -1 = 1 + log a

=> – 2 = log a

=> a = e-2

From (1), we have

Now, b = e-2 (-2) = -2 e-2

Given, (x1, y1) = (e-2, -2 e-2)

The equation of normal is,

y + 2/e2 = 1(x – 1/e2)

y + 2/e2 = x – 1/e2

x – y = 3/e2

### (i) which is parallel to the line 2x − y + 9 = 0?

Solution:

We have, y = x2 − 2x + 7

On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 2x – y + 9 = 0

So the slope of line is 2.

According to the question,

=> 2x – 2 = 2

=> 2x = 4

=> x = 2

=> y = 22 − 2(2) + 7 = 4 – 4 + 7 = 7

As (x1, y1) is (2, 7), the equation of tangent is,

y – 7 = 2(x – 2)

y – 7 = 2x – 4

y – 2x – 3 = 0

### (ii) which is perpendicular to the line 5y − 15x = 13.

Solution:

We have, y = x2 − 2x + 7

On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 5y − 15x = 13.

=> y = 3x + 13/5

So the slope of line is 3.

According to the question,

=> 2x – 2= -1/3

=> 6x – 6 = -1

=> x = 5/6

And y = 217/36.

As (x1, y1) is (5/6, 217/36), the equation of tangent is,

y – y1 = m (x – x1)

y – 217/36 = (-1/3) (x – 5/6)

36y -217 = -12x + 10

36y + 12x – 227 = 0

### Question 14. Find the equations of all lines having slope 2 and that are tangent to the curve y = 1/x – 3, x ≠ 3.

Solution:

Let (a , b) be the point where the tangent is drawn to this curve.

Since, the point lies on the curve, hence b = 1/(a – 3)

Slope of tangent, m =

Slope of the tangent} = 2

(a – 3)2 = – 2

a – 3 = √-2, which does not exist because 2 is negative.

So, there does not exist any such tangent.

### Question 15. Find the equations of all lines of slope zero and that are tangent to the curve .

Solution:

Slope of the given tangent is 0.

Let (a, b) be a point where the tangent is drawn to the curve.

Since, the point lies on the curve, hence b = . . . . (1)

Slope of tangent =

Given that, slope of tangent = slope of the given line,

=> -2 a + 2 = 0

=> 2a = 2

=> a = 1

From (1), we get

Now, b =

(a, b) = (1, 1/2)

The equation of tangent is,

y – y1 = m (x – x1)

y – 1/2 = 0 (x – 1)

y = 1/2

### Question 16. Find the equation of the tangent to the curve  which is parallel to the 4x − 2y + 5 = 0.

Solution:

We have,

Let (a, b) be the point where the tangent is drawn to the curve y =

On differentiating both sides, we get

Slope of tangent at (a, b) =

Slope of line 4x − 2y + 5 = 0 is 2.

Given that, slope of tangent = slope of the given line

9 = 16 (3a – 2)

9/16  = 3a – 2

3a = 9/16 + 2

a = 41/48

Now, b =

Therefore, (a, b) = (41/48, 3/4)

The equation of tangent is,

y – y1 = m (x – x1)

y – 3/4 = 2 (x – 41/48)

(4y – 3)/4 = 2 (48x – 41)/48

24y – 18 = 48x – 41

48x – 24y – 23 = 0

### Question 17. Find the equation of the tangent to the curve x2 + 3y − 3 = 0, which is parallel to the line y= 4x − 5.

Solution:

Suppose (a, b) be the required point.

We can find the slope of the given line by differentiating the equation w.r.t  x,

So, slope of the line = 4

Since (a, b) lies on the curve, we get a2 + 3b − 3 = 0  . . . . (1)

Now,

2x + 3dy/dx = 0

dy/dx = -2x/3

Slope of tangent, m=

Given that tangent is parallel to the line, So we get,

Slope of tangent, m = slope of the given line

=> -2a/3 = 4

=> a = -6

From (1), we get

=> 36 + 3b – 3 = 0

=> 3b = – 33

=> b = – 11

(a, b) = (-6, -11)

The equation of tangent is,

y – y1 = m (x – x1)

y + 11 = 4 (x + 6)

y + 11 = 4x + 24

4x – y + 13 = 0

### Question 18. Prove that  touches the straight line  for all n ∈ N, at the point (a, b) ?

Solution:

We have,

On differentiating both sides, we get

Slope of tangent =

The equation of tangent is,

y – b = -b/a (x – a)

ya – ab = – xb + ab

xb + ya = 2ab

So, the given line touches the given curve at the given point.

Hence proved.

### Question 19. Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = π/4.

Solution:

We have,

x = sin 3t, y = cos 2t

dx/dt = 3 cos3t and dy/dt = -2sin2t

Slope of tangent, m=

x1 = sin 3π/4 = 1/√2 and y1 = cos π/2 = 0

So, (x1, y1) = (1/√2, 0)

The equation of tangent is,

y – y1 = m (x – x1)

3y = 2√2 x – 2

2√2 x – 3y – 2 = 0

### Question 20. At what points will be tangents to the curve y = 2x3 − 15x2 + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points?

Solution:

We have,

y = 2x3 − 15x2 + 36x − 21

The slope of x – axis is 0.

Let (a, b) be the required point.

Since (a, b) lies on the curve, we get

b = 2a3 − 15a2 + 36a − 21   . . . . (1)

Also, we have

dy/dx = 6 x2 – 30x + 36

Slope of tangent at (a, b) =

Given that the slope of the tangent = slope of the x-axis, we have

=> 6a2 – 30a + 36 = 0

=> a2 – 5a + 6 = 0

=> (a – 2) (a – 3) = 0

=> a = 2 or a= 3

=> b = 7 or 6

When a = 2 and b = 7, the equation is,

y – y1 = m (x – x1)

y – 7 = 0 (x – 2)

y = 7

When a = 3 and b = 6, the equation is,

y – y1 = m (x – x1)

y – 6 = 0 (x – 3)

y = 6

### Question 21. Find the equation of the tangents to the curve 3x2 – y2 = 8, which passes through the point (4/3, 0).

Solution:

We have,

3x2 – y2 = 8  . . . . (1)

On differentiating both sides w.r.t x, we get

6x – 2y dy/dx = 0

2y dy/dx = 6x

dy/dx = 6x/2y

dy/dx = 3x/y

Let tangent at (h, k) pass through (4/3, 0). Since, (h, k) lies on (1), we get

3 h2 – k2 = 8 . . . (ii)

Slope of tangent at (h, k) = 3h/k

The equation of tangent at (h, k) is given by,

y – k = 3h/k (x – h)

Also,

=> 0 – k = (3h/k) (4/3 – h)

=> -k = 4h/k – 3h2/k

=> – k2 = 4h – 3h2

=> 8 – 3 h2 = 4h – 3 h2

=> 8 = 4h

=> h = 2

Also we get,

=> 12 – k2 = 8

=> k2 = 4

=> k = ±2

So, the points on curve (i) at which tangents pass through (4/3, 0) are (2, ±2).

When h = 2 and k = 2, the equation is,

y – 2 = (6/2) (x – 2)

3x – y – 4 = 0

When h = 2 and k = –2, the equation is,

y + 2 = (6/-2) (x – 2)

3x + y – 4 = 0

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