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

## Chapter 16 - Tangents and Normals Exercise MCQ

The abscissa of the point on the curve the normal at which passes through the origin is

a. 1

b.

c. 2

d.

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.

Two curves

a. touch each other

b. cut at right angle

c. cut at an angle

d. cut at an angle

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.

The tangent to the curve makes with x-axis an angle

a. 0

b.

c.

d.

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.

The tangent to the curve at the point (0, 1) meets x-axis at:

a. (0, 1)

b.

c. (2, 0)

d. (0, 2)

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

The equation of tangent to the curve where it crosses x-axis is

a. x + 5y = 2

b. x - 5y = 2

c. 5x - y = 2

d. 5x + y = 2

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

The points at which the tangents to the curve are parallel to x-axis are

a. (2, -2) (-2, -34)

b. (2, 34) (-2, 0)

c. (0, 34) (-2, 0)

d. (2, 2) (-2, 34)

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).

The curve has at (0, 0)

a. a vertical tangent

b. a horizontal tangent

c. an oblique tangent

d. no tangent

Given curve is

At (0, 0), we have

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

The equation to the normal to the curve y = sin x at (0,0) is

- x = 0
- y = 0
- x + y = 0
- x - y = 0

Correct option: (c)

The equation of the normal to the curve y = x + sin x cos x at x = π/2 is

- x = 2
- x = π
- x + π = 0
- 2x = π

Correct option: (d)

The equation of the normal to the curve y = x (2-x) at the point (2,0) is

- x - 2y = 2
- x - 2y + 2 = 0
- 2x + y = 4
- 2x + y - 4 = 0

Correct option: (a)

The point on the curve y^{2} = x where tangent makes 45° angles with x-axis is

- (1/2,1/4)
- (1/4,1/2)
- (4,2)
- (1,1)

Correct option: (b)

If the tangent to the curve x = at^{2}, y=2at is perpendicular to x-axis , then its point of contact is

- (a, a)
- (0, a)
- (0, 0)
- (a, 0)

Correct option: (c)

The point on the curve y = x^{2} - 3x + 2 where tangent is perpendicular to y = x is

- (0,2)
- (1,0)
- (-1,6)
- (2,-2)

Correct option: (b)

The point on the curve y^{2} = x where tangent makes 45° angle with x-axis is

- (1/2,1/4)
- (1/4,1/2)
- (4,2)
- (1,1)

Correct option:(b)

The point on the curve y = 12x - x^{2} where the slope of the tangent is zero will be

- (0,0)
- (2,16)
- (3,9)
- (6,36)

Correct option: (d)

The angle between the curves y^{2} = x and x^{2} = y at (1,1) is

Correct option: (b)

The equation of the normal to the curve 3x^{2} - y^{2} = 8 which is parallel to x + 3y = 8 is

- x - 3y = 8
- x - 3y + 8 = 0
- x + 3y ± 8 = 0
- x + 3y = 0

Correct option: (c)

The equation of tangent at those points where the curve y = x^{2} - 3x + 2 meets x-axis are

- x - y + 2 = 0 = x - y - 1
- x + y - 1 = 0 = x - y - 2
- x - y - 1 = 0 = x - y
- x - y = 0 = x + y

Correct option: (b)

The slope of tangent to the curve x = t^{2} + 3t - 8, y = 2t^{2} - 2t - 5 at point (2,-1) is

- 22 /7
- 6/7
- -6
- 7/6

Correct option: (b)

The what points the slope of the tangent to the curve x^{2} + y^{2} - 2x - 3 = 0 is zero

- (3,0), (-1,0)
- (3,0), (1,2)
- (-1,0) , (1,2)
- (1,2), (1,-2)

Correct option: (d)

The angle of intersection of the curves xy = a^{2} and x^{2} - y^{2} = 2a^{2} is

- 0°
- 45°
- 90°
- 30°

Correct option: (c)

If the curve ay + x^{2} = 7 and x^{3} = y cut orthogonally at (1,1), then a is equal to

- 1
- -6
- 6
- 0

Correct option: (c)

If the line y = x touches the curve y = x^{2} + bx + c at a point (1,1) then

- b = 1, c = 2
- b =-1, c = 1
- b = 2, c = 1
- b = -2, c = 1

Correct option: (b)

The slope of the tangent to the curve x = 3t^{2} + 1, y = t^{3}-1 at x = 1 is

- 1/2
- 0
- -2
- ∞

Correct option: (b)

The curves y = ae^{x} and y = be^{-x} cut orthogonally, if

- a = b
- a = -b
- ab = 1
- ab =2

Correct option: (c)

The equation of the normal to the curve x = a cos^{3}θ, y = a sin^{3}θ at the point θ = π/4 is

- x = 0
- y = 0
- x = y
- x + y = a

Correct option: (c)

If the curves y = 2 e^{x} and y = ae^{-x} intersect orthogonally, then a =

- 1/2
- -1/2
- 2
- 2e
^{2}

Correct option: (a)

The point on the curve y = 6x - x^{2} at which the tangent to the curve is inclined at π/4 to the line x + y = 0

- (-3,-27)
- (3,9)
- (7/2, 35/4)
- (0,0)

Correct option: (b)

The angle of intersection of the parabolas y^{2} = 4 ax and x^{2} = 4ay at the origin is

- π/6
- π/3
- π/2
- π/6

Correct option: (c)

The angle of intersection of the curve y = 2 sin^{2} x and y = cos 2 x at x

- π/4
- π /2
- π /3
- π /6

Correct option: (c)

Any tangent to the curve y = 2x^{7} + 3x + 5

- is parallel to x-axis
- is parallel to y -axis
- makes an acute angle with x- axis
- makes an obtuse angle with x -axis

Correct option: (c)

The point on the curve 9y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axis is

- (4, ±8/3)
- (-4, 8/3)
- (-4,-8/3)
- (8/3,4)

Correct option: (a)

The slope of the tangent to the curve x = t^{2} + 3t - 8, y = 2t^{2} - 2t - 5 at the point (2,-1) is

- 22/7
- 6/7
- 7/6
- -6/7

Correct option: (b)

The line y = mx + 1 is a tangent to the curve y^{2} = 4x, if the value of m is

- 1
- 2
- 3
- 1/2

Correct option: (a)

The normal at the point (1,1) on the curve 2y + x^{2} = 3 is

- x + y = 0
- x - y = 0
- x + y +1 = 0
- x - y = 1

Correct option: (b)

The normal to the curve x^{2} = 4y passing through (1,2) is

- x + y = 3
- x - y = 3
- x + y = 1
- x - y = 1

NOTE: Options are incorrect.

## Chapter 16 - Tangents and Normals Exercise Ex. 16.1

Find the slopes of the tangent and the normal to the curve x = a(θ - sinθ), y =a(1 + cos θ) at θ = .

## Chapter 16 - Tangents and Normals Exercise Ex. 16.2

Find the equations of the tangent and normal to the given curves at the indicated points:

y=x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (x = 1)

Find the equation of the normal to curve y^{2} = 4x at the point (1, 2) and also find the tangent.

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

Find the equations of the tangent and the normal to the given curves at the indicated points:

From (A)

Equation of tangent is

Find the equations of the tangent and the normal to the following curves at the indicated points:

X = 3 cosθ
- cos^{3}θ , y = 3 sinθ
- sin^{3}θ

Find the equation of the tangents to
the curve
3x^{2} - y^{2} = 8, which passes through the point (4/3, 0).

Find the equations of the tangent and the normal to the following curves at the indicated points:

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

Find the equations of the tangent and the normal to the following curves at the indicated points:

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

At what points will tangents to the curve be parallel to x-axis? Also, find the equations of the tangents to the curve at these points.

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

## Chapter 16 - Tangents and Normals Exercise Ex. 16.3

Find the angle of intersection of the folloing curves

Find the angle of intersection of the following curves:

Y = 4 - x^{2} and y = x^{2}

Prove that the curves xy =
4 and x^{2} + y^{2} = 8 touch each other.

Prove that the curves y^{2} = 4x and x^{2}
+ y^{2} - 6x + 1 = 0 touch each other at the point (1, 2)

## Chapter 16 - Tangents and Normals Exercise Ex. 16VSAQ

Find the slope of the tangent to the curve x = t^{2} + 3t - 8, y = 2t^{2} - 2t - 5 at t = 2.

^{}

Write the equation of the tangent drawn to the curve y = sin x at the point (0, 0).

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

Find the slope of the tangent to the curve at

Given curve is

Slope of tangent at (0, 0) is

Hence, slope of tangent at is 0.

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