RD Sharma Solutions for Class 12 Maths Chapter 16–Tangents and Normals

Class 12: Maths Chapter 16 solutions. Complete Class 12 Maths Chapter 16 Notes.

RD Sharma Solutions for Class 12 Maths Chapter 16–Tangents and Normals

RD Sharma 12th Maths Chapter 16, Class 12 Maths Chapter 16 solutions

Exercise 16.1 Page No: 16.10

1. Find the Slopes of the tangent and the normal to the following curves at the indicated points:

Solution:

(ii) y = √x at x = 9

Solution:

⇒ The Slope of the normal = – 6

(iii) y = x³ – x at x = 2

Solution:

(iv) y = 2x² + 3 sin x at x = 0

Solution:

(v) x = a (θ – sin θ), y = a (1 + cos θ) at θ = -π /2

Solution:

(vi) x = a cos³ θ, y = a sin³ θ at θ = π /4

Solution:

(vii) x = a (θ – sin θ), y = a (1 – cos θ) at θ = π /2

Solution:

(viii) y = (sin 2x + cot x + 2)² at x = π /2

Solution:

(ix) x² + 3y + y² = 5 at (1, 1)

Solution:

(x) x y = 6 at (1, 6)

Solution:

2. Find the values of a and b if the Slope of the tangent to the curve x y + a x + by = 2 at (1, 1) is 2.

Solution:

⇒ – a – 1 = 2(1 + b)

⇒ – a – 1 = 2 + 2b

⇒ a + 2b = – 3 … (1)

Also, the point (1, 1) lies on the curve xy + ax + by = 2, we have

1 × 1 + a × 1 + b × 1 = 2

⇒ 1 + a + b = 2

⇒ a + b = 1 … (2)

From (1) & (2), we get b = -4

Substitute b = – 4 in a + b = 1

a – 4 = 1

⇒ a = 5

So the value of a = 5 & b = – 4

3. If the tangent to the curve y = x³ + a x + b at (1, – 6) is parallel to the line x – y + 5 = 0, find a and b

Solution:

The given line is x – y + 5 = 0

y = x + 5 is the form of equation of a straight line y = mx + c, where m is the Slope of the line.

So the slope of the line is y = 1 × x + 5

So the Slope is 1. … (2)

Also the point (1, – 6) lie on the tangent, so

x = 1 & y = – 6 satisfies the equation, y = x³ + ax + b

– 6 = 1³ + a × 1 + b

⇒ – 6 = 1 + a + b

⇒ a + b = – 7 … (3)

Since, the tangent is parallel to the line, from (1) & (2)

Hence, 3 + a = 1

⇒ a = – 2

From (3)

a + b = – 7

⇒ – 2 + b = – 7

⇒ b = – 5

So the value is a = – 2 & b = – 5

4. Find a point on the curve y = x³ – 3x where the tangent is parallel to the chord joining (1, – 2) and (2, 2).

Solution:

5. Find a point on the curve y = x³ – 2x² – 2x at which the tangent lines are parallel to the line y = 2x – 3.

Solution:

Given the curve y = x³ – 2x² – 2x and a line y = 2x – 3

First, we will find the slope of tangent

y = x³ – 2x² – 2x

y = 2x – 3 is the form of equation of a straight line y = mx + c, where m is the Slope of the line.

So the slope of the line is y = 2 × (x) – 3

Thus, the Slope = 2. … (2)

From (1) & (2)

⇒ 3x² – 4x – 2 = 2

⇒ 3x² – 4x = 4

⇒ 3x² – 4x – 4 = 0

We will use factorization method to solve the above Quadratic equation.

⇒ 3x² – 6x + 2x – 4 = 0

⇒ 3 x (x – 2) + 2 (x – 2) = 0

⇒ (x – 2) (3x + 2) = 0

⇒ (x – 2) = 0 & (3x + 2) = 0

⇒ x = 2 or

x = -2/3

Substitute x = 2 & x = -2/3 in y = x³ – 2x² – 2x

When x = 2

⇒ y = (2)³ – 2 × (2)² – 2 × (2)

⇒ y = 8 – (2 × 4) – 4

⇒ y = 8 – 8 – 4

⇒ y = – 4

6. Find a point on the curve y² = 2x³ at which the Slope of the tangent is 3

Solution:

Given the curve y² = 2x³ and the Slope of tangent is 3

y² = 2x³

Differentiating the above with respect to x

dy/dx = 0 which is not possible.

So we take x = 2 and substitute it in y² = 2x³, we get

y² = 2(2)³

y² = 2 × 8

y² = 16

y = 4

Thus, the required point is (2, 4)

7. Find a point on the curve x y + 4 = 0 at which the tangents are inclined at an angle of 45^o with the x–axis.

Solution:

Substitute in xy + 4 = 0, we get

⇒ x (– x) + 4 = 0

⇒ – x² + 4 = 0

⇒ x² = 4

⇒ x =
2

So when x = 2, y = – 2

And when x = – 2, y = 2

Thus, the points are (2, – 2) & (– 2, 2)

8. Find a point on the curve y = x² where the Slope of the tangent is equal to the x – coordinate of the point.

Solution:

Given the curve is y = x²

y = x²

Differentiating the above with respect to x

From (1) & (2), we get,

2x = x

⇒ x = 0.

Substituting this in y = x², we get,

y = 0²

⇒ y = 0

Thus, the required point is (0, 0)

9. At what point on the circle x² + y² – 2x – 4y + 1 = 0, the tangent is parallel to x – axis.

Solution:

⇒ – (x – 1) = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0, we get,

⇒ 1² + y² – 2(1) – 4y + 1 = 0

⇒ 1 – y² – 2 – 4y + 1 = 0

⇒ y² – 4y = 0

⇒ y (y – 4) = 0

⇒ y = 0 and y = 4

Thus, the required point is (1, 0) and (1, 4)

10. At what point of the curve y = x² does the tangent make an angle of 45^o with the x–axis?

Solution:

Given the curve is y = x²

Differentiating the above with respect to x

⇒ y = x²

Exercise 16.2 Page No: 15.27

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

Solution:

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

Solution:

3. Find the equation of the tangent and the normal to the following curves at the indicated points:

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)

Solution:

(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at x = 1 y = 3

Solution:

(iii) y = x² at (0, 0)

Solution:

Given y = x² at (0, 0)

By differentiating the given curve, we get the slope of the tangent

m (tangent) at (x = 0) = 0

Normal is perpendicular to tangent so, m₁m₂ = – 1

We can see that the slope of normal is not defined

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

y = 0

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

x = 0

(iv) y = 2x² – 3x – 1 at (1, – 2)

Solution:

Given y = 2x² – 3x – 1 at (1, – 2)

By differentiating the given curve, we get the slope of the tangent

m (tangent) at (1, – 2) = 1

Normal is perpendicular to tangent so, m₁m₂ = – 1

m (normal) at (1, – 2) = – 1

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

y + 2 = 1(x – 1)

y = x – 3

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

y + 2 = – 1(x – 1)

y + x + 1 = 0

Solution:

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

y + 2 = – 2(x – 2)

y + 2x = 2

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

2y + 4 = x – 2

2y – x + 6 = 0

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

Solution:

5. Find the equation of the tangent and the normal to the following curves at the indicated points:

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

Solution:

Given x = θ + sin θ, y = 1 + cos θ at θ = π/2

By differentiating the given equation with respect to θ, we get the slope of the tangent

Solution:

(iii) x = at², y = 2at at t = 1.

Solution:

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

Solution:

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

Solution:

(vi) x = 3 cos θ – cos³ θ, y = 3 sin θ – sin³θ

Solution:

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

Solution:

Finding y co – ordinate by substituting x in the given curve

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

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

x = 2

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

Solution:

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:

Given y² = ax³ + b is y = 4x – 5

By differentiating the given curve, we get the slope of the tangent

m (tangent) at (2, 3) = 2a

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

Now comparing 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

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:

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

Solution:

Exercise 16.3 Page No: 16.40

1. Find the angle to intersection of the following curves:

(i) y² = x and x² = y

Solution:

x = y²

When y = 0, x = 0

When y = 1, x = 1

Substituting above values for m₁ & m_2, we get,

When x = 0,

(ii) y = x² and x² + y² = 20

Solution:

(iii) 2y² = x³ and y² = 32x

Solution:

Substituting (2) in (1), we get

⇒ 2y² = x³

⇒ 2(32x) = x³

⇒ 64 x = x³

⇒ x³ – 64x = 0

⇒ x (x² – 64) = 0

⇒ x = 0 & (x² – 64) = 0

⇒ x = 0 & ±8

Substituting x = 0 & x = ±8 in (1) in (2),

y² = 32x

When x = 0, y = 0

When x = 8

⇒ y² = 32 × 8

⇒ y² = 256

⇒ y = ±16

Substituting above values for m₁ & m_2, we get,

When x = 0, y = 16

(iv) x² + y² – 4x – 1 = 0 and x² + y² – 2y – 9 = 0

Solution:

Given curves x² + y² – 4x – 1 = 0 … (1) and x² + y² – 2y – 9 = 0 … (2)

First curve is x² + y² – 4x – 1 = 0

⇒ x² – 4x + 4 + y² – 4 – 1 = 0

⇒ (x – 2)² + y² – 5 = 0

Now, Subtracting (2) from (1), we get

⇒ x² + y² – 4x – 1 – ( x² + y² – 2y – 9) = 0

⇒ x² + y² – 4x – 1 – x² – y² + 2y + 9 = 0

⇒ – 4x – 1 + 2y + 9 = 0

⇒ – 4x + 2y + 8 = 0

⇒ 2y = 4x – 8

⇒ y = 2x – 4

Substituting y = 2x – 4 in (3), we get,

⇒ (x – 2)² + (2x – 4)² – 5 = 0

⇒ (x – 2)² + 4(x – 2)² – 5 = 0

⇒ (x – 2)²(1 + 4) – 5 = 0

⇒ 5(x – 2)² – 5 = 0

⇒ (x – 2)² – 1 = 0

⇒ (x – 2)² = 1

⇒ (x – 2) = ±1

⇒ x = 1 + 2 or x = – 1 + 2

⇒ x = 3 or x = 1

So, when x = 3

y = 2×3 – 4

⇒ y = 6 – 4 = 2

So, when x = 3

y = 2 × 1 – 4

⇒ y = 2 – 4 = – 2

The point of intersection of two curves are (3, 2) & (1, – 2)

Solution:

2. Show that the following set of curves intersect orthogonally:
(i) y = x³ and 6y = 7 – x²

Solution:

Given curves y = x³ … (1) and 6y = 7 – x² … (2)

Solving (1) & (2), we get

⇒ 6y = 7 – x²

⇒ 6(x³) = 7 – x²

⇒ 6x³ + x² – 7 = 0

Since f(x) = 6x³ + x² – 7,

We have to find f(x) = 0, so that x is a factor of f(x).

When x = 1

f (1) = 6(1)³ + (1)² – 7

f (1) = 6 + 1 – 7

f (1) = 0

Hence, x = 1 is a factor of f(x).

Substituting x = 1 in y = x³, we get

y = 1³

y = 1

The point of intersection of two curves is (1, 1)

First curve y = x³

Differentiating above with respect to x,

(ii) x³ – 3xy² = – 2 and 3x² y – y³ = 2

Solution:

Given curves x³ – 3xy² = – 2 … (1) and 3x²y – y³ = 2 … (2)

Adding (1) & (2), we get

⇒ x³ – 3xy² + 3x²y – y³ = – 2 + 2

⇒ x³ – 3xy² + 3x²y – y³ = – 0

⇒ (x – y)³ = 0

⇒ (x – y) = 0

⇒ x = y

Substituting x = y on x³ – 3xy² = – 2

⇒ x³ – 3 × x × x² = – 2

⇒ x³ – 3x³ = – 2

⇒ – 2x³ = – 2

⇒ x³ = 1

⇒ x = 1

Since x = y

y = 1

The point of intersection of two curves is (1, 1)

First curve x³ – 3xy² = – 2

Differentiating above with respect to x,

(iii) x² + 4y² = 8 and x² – 2y² = 4.

Solution:

3. x² = 4y and 4y + x² = 8 at (2, 1)

Solution:

(ii) x² = y and x³ + 6y = 7 at (1, 1)

Solution:

Given curves x² = y … (1) and x³ + 6y = 7 … (2)

The point of intersection of two curves (1, 1)

Solving (1) & (2), we get,

First curve is x² = y

Differentiating above with respect to x,

(iii) y² = 8x and 2x² + y² = 10 at (1, 2√2)

Solution:

Given curves y² = 8x … (1) and 2x² + y² = 10 … (2)

4. Show that the curves 4x = y² and 4xy = k cut at right angles, if k² = 512.

Solution:

5. Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8.

Solution:

Given curves 2x = y² … (1) and 2xy = k … (2)

We have to prove that two curves cut at right angles if k² = 8

Now, differentiating curves (1) & (2) with respect to x, we get

⇒ 2x = y²

RD Sharma Solutions for Class 12 Maths Chapter 16: Download PDF

RD Sharma Solutions for Class 12 Maths Chapter 16–Tangents and Normals

Download PDF: RD Sharma Solutions for Class 12 Maths Chapter 16–Tangents and Normals PDF

Chapterwise RD Sharma Solutions for Class 12 Maths :

• Chapter 1–Relation
• Chapter 2–Functions
• Chapter 3–Binary Operations
• Chapter 4–Inverse Trigonometric Functions
• Chapter 5–Algebra of Matrices
• Chapter 6–Determinants
• Chapter 7–Adjoint and Inverse of a Matrix
• Chapter 8–Solution of Simultaneous Linear Equations
• Chapter 9–Continuity
• Chapter 10–Differentiability
• Chapter 11–Differentiation
• Chapter 12–Higher Order Derivatives
• Chapter 13–Derivatives as a Rate Measurer
• Chapter 14–Differentials, Errors and Approximations
• Chapter 15–Mean Value Theorems
• Chapter 16–Tangents and Normals
• Chapter 17–Increasing and Decreasing Functions
• Chapter 18–Maxima and Minima
• Chapter 19–Indefinite Integrals

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.