# Class 12-science RD SHARMA Solutions Maths Chapter 16 - Tangents and Normals

## Tangents and Normals Exercise MCQ

### Solution 30

Let (p, q) be the point on the given curve at which the normal passes through the origin.

Therefore,

Now, the equation of normal at (p, q) is passing through origin.

Therefore,

As p = 1 satisfies the equation, therefore the abscissa is 1.

### Solution 31

The curves are

… (i)

… (ii)

Differentiating (i) w.r.t x, we get

Differentiating (ii) w.r.t x, we get

Now,

So, both the curves are cut each other at right angle.

### Solution 32

Given:

Differentiating 'x' and 'y' w.r.t t, we get

Dividing (ii) by (i), we get

Hence, the tangent to the curve makes an angle of with x-axis.

### Solution 33

Given curve is

Therefore, slope of tangent is 2.

The equation of tangent is y - 1 = 2(x - 0)

i.e. y = 2x + 1

This equation of tangent meets x-axis when y = 0

Thus, the required point is

### Solution 34

Given curve is

The curve crosses x-axis when y = 0

Therefore, x = 2

So, the tangent touches the curve at point (2, 0).

The equation of tangent at (2, 0) is

### Solution 35

Given curve is

As the tangents are parallel to x-axis, their slope will be 0.

When x = 2, y = 2^{3}
- 12 × 2 + 18 = 2

When x = -2, y =
(-2)^{3} - 12(-2) + 18 = 34

So, the points are (2, 2) and (-2, 34).

### Solution 36

Given curve is

At (0, 0), we have

Thus, the given curve has vertical tangent, which is parallel to y-axis, at (0, 0).

### Solution 1

Correct option: (c)

### Solution 2

Correct option: (d)

### Solution 3

Correct option: (a)

### Solution 4

Correct option: (b)

### Solution 5

Correct option: (c)

### Solution 6

Correct option: (b)

### Solution 7

Correct option:(b)

### Solution 8

Correct option: (d)

### Solution 9

Correct option: (b)

### Solution 10

Correct option: (c)

### Solution 11

Correct option: (b)

### Solution 12

Correct option: (b)

### Solution 13

Correct option: (d)

### Solution 14

Correct option: (c)

### Solution 15

Correct option: (c)

### Solution 16

Correct option: (b)

### Solution 17

Correct option: (b)

### Solution 18

Correct option: (c)

### Solution 19

Correct option: (c)

### Solution 20

Correct option: (a)

### Solution 21

Correct option: (b)

### Solution 22

Correct option: (c)

### Solution 23

Correct option: (c)

### Solution 24

Correct option: (c)

### Solution 25

Correct option: (a)

### Solution 26

Correct option: (b)

### Solution 27

Correct option: (a)

### Solution 28

Correct option: (b)

### Solution 29

NOTE: Options are incorrect.

## Tangents and Normals Exercise Ex. 16.1

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 1(vii)

### Solution 1(viii)

### Solution 1(ix)

### Solution 1(x)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

## Tangents and Normals Exercise Ex. 16.2

### Solution 1

### Solution 2

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 3(v)

### Solution 3(vi)

### Solution 3(vii)

### Solution 3(viii)

### Solution 3(ix)

### Solution 3(x)

### Solution 3(xi)

### Solution 3(xii)

### Solution 3(xiii)

### Solution 3(xiv)

### Solution 3(xv)

### Solution 3(xvi)

The equation of the given curve is y^{2} = 4x . Differentiating with respect to x, we have:

### Solution 3(xix)

### Solution 4

### Solution 5(i)

### Solution 5(ii)

From (A)

Equation of tangent is

### Solution 5(iii)

### Solution 5(iv)

### Solution 5(v)

### Solution 5(vi)

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 21

### Solution 3(xvii)

Given equation curve is

Differentiating w.r.t x, we get

Slope of tangent at is

Slope of normal will be

Equation of tangent at will be

Equation of normal at is

### Solution 3(xviii)

Given equation curve is

Differentiating w.r.t x, we get

Slope of tangent at is

Slope of normal will be

Equation of tangent at will be

Equation of normal at is

### Solution 20

Given equation curve is

Differentiating w.r.t x, we get

As tangent is parallel to x-axis, its slope will be m = 0

As this point lies on the curve, we can find y

Or

So, the points are (3, 6) and (2, 7).

Equation of tangent at (3, 6) is

y - 6 = 0 (x - 3)

y - 6 = 0

Equation of tangent at (2, 7) is

y - 7 = 0 (x - 2)

y - 7 = 0

## Tangents and Normals Exercise Ex. 16.3

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 1(vii)

### Solution 1(viii)

### Solution 1 (ix)

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8(i)

### Solution 8(ii)

### Solution 9

### Solution 10

## Tangents and Normals Exercise Ex. 16VSAQ

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

Given curve is y = sin x

Slope of tangent at (0, 0) is

So, the equation of tangent at (0, 0) is

y - 0 = 1 (x - 0)

y = x

### Solution 19

Given curve is

Slope of tangent at (0, 0) is

Hence, slope of tangent at is 0.