RD Sharma Solutions for Class 12 Maths Exercise 16.3 Chapter 16 Tangents and Normals

RD Sharma Solutions for Class 12 Maths Exercise 16.3 Chapter 16 Tangents and Normals are provided here. RD Sharma Solutions for Class 12 are available on the BYJU’S website in PDF format with solutions prepared by experienced faculty. This exercise explains the angle of intersection of two curves. If we draw tangents to these curves at the intersecting point, the angle between these tangents is called the angle between two curves.

The PDF of RD Sharma Solutions for Class 12 Maths Chapter 16 Tangents and Normals Exercise 16.3 are provided here. Some of the important topics of this exercise are listed below.

• Finding the equations of tangent and normal
• Finding the equation of the curve
• The angle of intersection of two curves
• Orthogonal curves

RD Sharma Solutions for Class 12 Tangents and Normals Exercise 16.3:

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Exercise 16.3 Page No: 16.40

1. Find the angle of 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