# NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT solutions for exercise 6.3 Class 12 Maths chapter 6 is related to the topic normal and tangents. The equations of normals and tangent to a curve and questions related to these are discussed in exercise 6.3 Class 12 Maths. There are twenty-seven questions presented in the Class 12 Maths chapter 6 exercise 6.3. These questions in Class 12th Maths chapter 6 exercise 6.3 are solved by our Mathematics subject matter experts and are reliable and according to the CBSE pattern. The NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 along with all other exercises of the NCERT chapter applications of derivatives gives a good idea of concepts discussed in the NCERT Book.

Application of Derivatives Exercise 6.1

Application of Derivatives Exercise 6.2

Application of Derivatives Exercise 6.4

Application of Derivatives Exercise 6.5

Application of Derivatives Miscellaneous Exercise

** ****Application of Derivatives Class 12 Chapter 6 Exercise 6.3**

**Application of Derivatives Class 12 Chapter 6 Exercise 6.3**

** Question:1 ** . Find the slope of the tangent to the curve

** Answer: **

Given curve is,

Now, the slope of the tangent at point x =4 is given by

** Question:2 ** . Find the slope of the tangent to the curve

** Answer: **

Given curve is,

The slope of the tangent at x = 10 is given by

at x = 10

hence, slope of tangent at x = 10 is

** Question:3 ** Find the slope of the tangent to curve at the point whose x-coordinate is 2.

** Answer: **

Given curve is,

The slope of the tangent at x = 2 is given by

Hence, the slope of the tangent at point x = 2 is 11

** Question:4 ** Find the slope of the tangent to the curve at the point whose x-coordinate is 3.

** Answer: **

Given curve is,

The slope of the tangent at x = 3 is given by

Hence, the slope of tangent at point x = 3 is 24

** Question:5 ** Find the slope of the normal to the curve

** Answer: **

The slope of the tangent at a point on a given curve is given by

Now,

Similarly,

Hence, the slope of the tangent at is -1

Now,

Slope of normal = =

Hence, the slope of normal at is 1

** Question:6 ** Find the slope of the normal to the curve

** Answer: **

The slope of the tangent at a point on given curves is given by

Now,

Similarly,

Hence, the slope of the tangent at is

Now,

Slope of normal = =

Hence, the slope of normal at is

** Question:7 ** Find points at which the tangent to the curve is parallel to the x-axis.

** Answer: **

We are given :

Differentiating the equation with respect to x, we get :

or

or

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

or

Thus, Either x = -1 or x = 3

When x = -1 we get y = 12 and if x =3 we get y = -20

So the required points are (-1, 12) and (3, -20).

** Question:8 ** Find a point on the curve at which the tangent is parallel to the chord joining the points (2, 0) and

(4, 4).

** Answer: **

Points joining the chord is (2,0) and (4,4)

Now, we know that the slope of the curve with given two points is

As it is given that the tangent is parallel to the chord, so their slopes are equal

i.e. slope of the tangent = slope of the chord

Given the equation of the curve is

Now, when

Hence, the coordinates are (3, 1)

** Question:9 ** Find the point on the curve at which the tangent is

** Answer: **

We know that the equation of a line is y = mx + c

Know the given equation of tangent is

y = x - 11

So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11

As we know that slope of the tangent at a point on the given curve is given by

Given the equation of curve is

When x = 2 ,

and

When x = -2 ,

Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is

** Question:10 ** Find the equation of all lines having slope –1 that are tangents to the curve

** Answer: **

We know that the slope of the tangent of at the point of the given curve is given by

Given the equation of curve is

It is given thta slope is -1

So,

Now, when x = 0 ,

and

when x = 2 ,

Hence, the coordinates are (0,-1) and (2,1)

Equation of line passing through (0,-1) and having slope = -1 is

y = mx + c

-1 = 0 X -1 + c

c = -1

Now equation of line is

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

Similarly, Equation of line passing through (2,1) and having slope = -1 is

y = mx + c

1 = -1 X 2 + c

c = 3

Now equation of line is

y = -x + 3 ** y + x - 3 = 0 **

** Question:11 ** Find the equation of all lines having slope 2 which are tangents to the curve

** Answer: **

We know that the slope of the tangent of at the point of the given curve is given by

Given the equation of curve is

It is given that slope is 2

So,

So, this is not possible as our coordinates are imaginary numbers

Hence, no tangent is possible with slope 2 to the curve

** Question:12 ** Find the equations of all lines having slope 0 which are tangent to the curve

** Answer: **

We know that the slope of the tangent at a point on the given curve is given by

Given the equation of the curve as

It is given thta slope is 0

So,

Now, when x = 1 ,

Hence, the coordinates are

Equation of line passing through and having slope = 0 is

y = mx + c

1/2 = 0 X 1 + c

c = 1/2

Now equation of line is

** Question:13(i) ** Find points on the curve at which the tangents are parallel to x-axis

** Answer: **

Parallel to x-axis means slope of tangent is 0

We know that slope of tangent at a given point on the given curve is given by

Given the equation of the curve is

From this, we can say that

Now. when ,

Hence, the coordinates are (0,4) and (0,-4)

** Question:13(ii) ** Find points on the curve at which the tangents are parallel to y-axis

** Answer: **

Parallel to y-axis means the slope of the tangent is , means the slope of normal is 0

We know that slope of the tangent at a given point on the given curve is given by

Given the equation of the curve is

Slope of normal =

From this we can say that y = 0

Now. when y = 0,

Hence, the coordinates are (3,0) and (-3,0)

** Question:14(i) ** Find the equations of the tangent and normal to the given curves at the indicated

points:

** Answer: **

We know that Slope of the tangent at a point on the given curve is given by

Given the equation of the curve

at point (0,5)

Hence slope of tangent is -10

Now we know that,

Now, equation of tangent at point (0,5) with slope = -10 is

equation of tangent is

Similarly, the equation of normal at point (0,5) with slope = 1/10 is

equation of normal is

** Question:14(ii) ** Find the equations of the tangent and normal to the given curves at the indicated

points:

** Answer: **

We know that Slope of tangent at a point on given curve is given by

Given equation of curve

at point (1,3)

Hence slope of tangent is 2

Now we know that,

Now, equation of tangent at point (1,3) with slope = 2 is

y = 2x + 1 ** y -2x = 1 **

Similarly, equation of normal at point (1,3) with slope = -1/2 is

y = mx + c

equation of normal is

** Question:14(iii) ** Find the equations of the tangent and normal to the given curves at the indicated

points:

** Answer: **

We know that Slope of the tangent at a point on the given curve is given by

Given the equation of the curve

at point (1,1)

Hence slope of tangent is 3

Now we know that,

Now, equation of tangent at point (1,1) with slope = 3 is

equation of tangent is

Similarly, equation of normal at point (1,1) with slope = -1/3 is

y = mx + c

equation of normal is

** Question:14(iv) ** Find the equations of the tangent and normal to the given curves at the indicated points

** Answer: **

We know that Slope of the tangent at a point on the given curve is given by

Given the equation of the curve

at point (0,0)

Hence slope of tangent is 0

Now we know that,

Now, equation of tangent at point (0,0) with slope = 0 is

y = 0

Similarly, equation of normal at point (0,0) with slope = is

** Question:14(v) ** Find the equations of the tangent and normal to the given curves at the indicated points:

** Answer: **

We know that Slope of the tangent at a point on the given curve is given by

Given the equation of the curve

Now,

and

Now,

Hence slope of the tangent is -1

Now we know that,

Now, the equation of the tangent at the point with slope = -1 is

and

equation of the tangent at

i.e. is

Similarly, the equation of normal at with slope = 1 is

and

equation of the tangent at

i.e. is

** Question:15(a) ** Find the equation of the tangent line to the curve which is parallel to the line

** Answer: **

Parellel to line means slope of tangent and slope of line is equal

We know that the equation of line is

y = mx + c

on comparing with the given equation we get slope of line m = 2 and c = 9

Now, we know that the slope of tangent at a given point to given curve is given by

Given equation of curve is

Now, when x = 2 ,

Hence, the coordinates are (2,7)

Now, equation of tangent paasing through (2,7) and with slope m = 2 is

y = mx + c

7 = 2 X 2 + c

c = 7 - 4 = 3

So,

y = 2 X x+ 3

y = 2x + 3

So, the equation of tangent is y - 2x = 3

** Question:15(b) ** Find the equation of the tangent line to the curve which is perpendicular to the line

** Answer: **

Perpendicular to line means

We know that the equation of the line is

y = mx + c

on comparing with the given equation we get the slope of line m = 3 and c = 13/5

Now, we know that the slope of the tangent at a given point to given curve is given by

Given the equation of curve is

Now, when ,

Hence, the coordinates are

Now, the equation of tangent passing through (2,7) and with slope is

So,

Hence, equation of tangent is 36y + 12x = 227

** Question:16 ** Show that the tangents to the curve at the points where x = 2 and x = – 2 are parallel .

** Answer: **

Slope of tangent =

When x = 2

When x = -2

Slope is equal when x= 2 and x = - 2

Hence, we can say that both the tangents to curve is parallel

** Question:17 ** Find the points on the curve at which the slope of the tangent is equal to the y-coordinate of the point.

** Answer: **

Given equation of curve is

Slope of tangent =

it is given that the slope of the tangent is equal to the y-coordinate of the point

We have

So, when x = 0 , y = 0

and when x = 3 ,

Hence, the coordinates are (3,27) and (0,0)

** Question:18 ** For the curve , find all the points at which the tangent passes

through the origin.

** Answer: **

Tangent passes through origin so, (x,y) = (0,0)

Given equtaion of curve is

Slope of tangent =

Now, equation of tangent is

at (0,0) Y = 0 and X = 0

and we have

Now, when x = 0,

when x = 1 ,

when x= -1 ,

Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

** Question:19 ** Find the points on the curve at which the tangents are parallel

to the x-axis.

** Answer: **

parellel to x-axis means slope is 0

Given equation of curve is

Slope of tangent =

When x = 1 ,

Hence, the coordinates are (1,2) and (1,-2)

** Question:20 ** Find the equation of the normal at the point for the curve

** Answer: **

Given equation of curve is

** Slope of tangent **

at point

Now, we know that

equation of normal at point and with slope

Hence, the equation of normal is

** Question:21 ** Find the equation of the normals to the curve which are parallel

to the line

** Answer: **

Equation of given curve is

Parellel to line means slope of normal and line is equal

We know that, equation of line

y= mx + c

on comparing it with our given equation. we get,

Slope of tangent =

We know that

Now, when x = 2,

and

When x = -2 ,

Hence, the coordinates are (2,18) and (-2,-6)

Now, the equation of at point (2,18) with slope

Similarly, the equation of at point (-2,-6) with slope

Hence, the equation of the normals to the curve which are parallel

to the line

are ** x +14y - 254 = 0 ** and ** x + 14y +86 = 0 **

** Question:22 ** Find the equations of the tangent and normal to the parabola at the point

** Answer: **

Equation of the given curve is

Slope of tangent =

at point

Now, the equation of tangent with point and slope is

We know that

Now, the equation of at point with slope -t

Hence, the equations of the tangent and normal to the parabola

at the point are

** Question:23 ** Prove that the curves and xy = k cut at right angles*

** Answer: **

Let suppose, Curve and xy = k cut at the right angle

then the slope of their tangent also cut at the right angle

means,

-(i)

Now these values in equation (i)

Hence proved

** Question:24 ** Find the equations of the tangent and normal to the hyperbola

at the point

** Answer: **

Given equation is

Now ,we know that

slope of tangent =

at point

equation of tangent at point with slope

Now, divide both sides by

** Hence, the equation of tangent is **

We know that

equation of normal at the point with slope

** Question:25 ** Find the equation of the tangent to the curve which is parallel to the line

** Answer: **

Parellel to line means the slope of tangent and slope of line is equal

We know that the equation of line is

y = mx + c

on comparing with the given equation we get the slope of line m = 2 and c = 5/2

Now, we know that the slope of the tangent at a given point to given curve is given by

Given the equation of curve is

Now, when

,

but y cannot be -ve so we take only positive value

Hence, the coordinates are

Now, equation of tangent paasing through

and with slope m = 2 is

Hence, equation of tangent paasing through and with slope m = 2 is 48x - 24y = 23

** Question:26 ** The slope of the normal to the curve is

(A) 3 (B) 1/3 (C) –3 (D) -1/3

** Answer: **

Equation of the given curve is

Slope of tangent =

at x = 0

Now, we know that

Hence, (D) is the correct option

** Question:27 ** The line is a tangent to the curve at the point

(A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)

** Answer: **

The slope of the given line is 1

given curve equation is

If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve

The slope of tangent =

Now, when y = 2,

Hence, the coordinates are (1,2)

Hence, (A) is the correct answer

**More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3**

As the questions in the Class 12 Maths chapter 6 exercise 6.3 deal with an application of derivatives, it is better for the students to revise the basic derivatives of trigonometric functions, exponential functions and some other special functions and rules and properties related to the derivatives. Out of the 27 problems in the Class 12th Maths chapter, 6 exercise 6. 3 question 26 is to find the slope of the normal to a given curve and question 27 asks to find the points for which a line is a tangent to the given curve.

**Also Read| **Application of Derivatives Class 12 Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3**

Solving exercise 6.3 Class 12 Maths will be beneficial for both the CBSE board exam and JEE Main exam.

One question from NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 can be expected for the CBSE Class 12 Maths board paper.

The applications of tangents will be used in Class 12 Physics and Chemistry also.

**Also see-**

NCERT Exemplar Solutions Class 12 Maths Chapter 6

NCERT Solutions for Class 12 Maths Chapter 6

**NCERT Solutions Subject Wise**

NCERT Solutions Class 12 Chemistry

NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Biology

NCERT Solutions for Class 12 Mathematics

**Subject Wise NCERT Exemplar Solutions**

NCERT Exemplar Class 12 Maths

NCERT Exemplar Class 12 Physics

NCERT Exemplar Class 12 Chemistry

NCERT Exemplar Class 12 Biology