# NCERT Solutions class 12 Maths Exercise 6.3 (Ex 6.3) Chapter 6 Application of Derivatives

## NCERT Solutions for Class 12 Maths Exercise 6.3 Chapter 6 Application of Derivatives – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 (Ex 6.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.3) Exercise 6.3

1. Find the slope of tangent to the curve  at

Ans. Given: Equation of the curve  ……….(i)

Slope of the tangent to the curve = Value of  at the point

Slope of the tangent at point  to the curve (i)

= 768 – 4 = 764

### 2. Find the slope of tangent to the curve  at

Ans. Given: Equation of the curve  ……….(i)

……….(ii)

Slope of the tangent at point  to the curve (i)

### 3. Find the slope of tangent to the curve  at the given point whose coordinate is 2.

Ans. Given: Equation of the curve  ……….(i)

Slope of the tangent at point  to the curve (i)

= 12 – 1 = 11

### 4. Find the slope of tangent to the curve  at the given point whose coordinate is 3.

Ans. Given: Equation of the curve ……….(i)

Slope of the tangent at point  to the curve (i)

= 27 – 3 = 24

### 5. Find the slope of the normal to the curve  at

Ans. Given: Equations of the curves are

and

and

and

Slope of the tangent at

And Slope of the normal at

= 1

### 6. Find the slope of the normal to the curve  at

Ans. Given: Equations of the curves are  and

and

and

Slope of the tangent at

And Slope of the normal at

### 7. Find the point at which the tangent to the curve  is parallel to the axis.

Ans. Given: Equation of the curve   ……….(i)

Since, the tangent is parallel to the axis, i.e.,

From eq. (i), when

when

Therefore, the required points are  and

### 8. Find the point on the curve  at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

Ans. Let the given points are A (2, 0) and B (4, 4).

Slope of the chord AB =

Equation of the curve is

Slope of the tangent at

If the tangent is parallel to the chord AB, then Slope of tangent = Slope of chord

Therefore, the required point is (3, 1).

### 9. Find the point on the curve  at which the tangent is

Ans. Given: Equation of the curve  ……….(i)

Equation of the tangent ……….(ii)

From eq. (i),

= Slope of the tangent at

But from eq. (ii), the slope of tangent =

From eq. (i), when

And when

Since  does not satisfy eq. (ii), therefore the required point is

### 10. Find the equation of all lines having slope  that are tangents to the curve

Ans. Given: Equation of the curve  ……….(i)

= Slope of the tangent at

But according to question, slope =

=

or

From eq. (i), when

And when

Points of contact are (2, 1) and

And Equation of two tangents are

and

=

### 11. Find the equations of all lines having slope 2 which are tangents to the curve

Ans. Given: Equation of the curve

= Slope of the tangent at

But according to question, slope =

= 2

which is not possible.

Hence, there is no tangent to the given curve having slope 2.

### 12. Find the equations of all lines having slope 0 which are tangents to the curve

Ans. Given: Equation of the curve  ……….(i)

But according to question, slope = 0

= 0

From eq. (i),

Therefore, the point on the curve which tangent has slope 0 is

Equation of the tangent is

### 13. Find the points on the curve  at which the tangents are:

(i) parallel to axis

(ii) parallel to axis

Ans. Given: Equation of the curve  ……….(i)

……….(ii)

(i) If tangent is parallel to axis, then Slope of tangent = 0

= 0

From eq. (i),

Therefore, the points on curve (i) where tangents are parallel to axis are .

(ii) If the tangent parallel to axis.

Slope of the tangent =

dxdy=0dxdy=0

From eq. (ii),

From eq. (i),

Therefore, the points on curve (i) where tangents are parallel to axis are .

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

(i)  at (0, 5)

(ii)  at (1, 3)

(iii)  at (1, 1)

(iv)  at (0, 0)

(v)     at

Ans. (i) Equation of the curve

slope of tangent  = at (0, 5)

= (say)

Slope of the normal at (0, 5) is

Equation of the tangent at (0, 5) is y5=10(x0)y−5=−10(x−0)

y5=10xy−5=−10x

And Equation of the normal at (0, 5) is

(ii) Equation of the curve

slope of tangent =   at (1, 3)

= (say)

Slope of the normal at (1, 3) is

Equation of the tangent at (1, 3) is

And Equation of the normal at (1, 3) is

(iii) Equation of the curve  ……….(i)

slope of tangent =   at (1, 1)

=     =  (say)

Slope of the normal at (1, 1) is

Equation of the tangent at (1, 1) is

And Equation of the normal at (1, 1) is

(iv) Equation of the curve  ……….(i)

slope of tangent =   at (0, 0)

=  (say)

Equation of the tangent at (0, 0) is

which is equaion of x-axis

thus    normal at (0, 0) is axis.

equation of normal is

x=0

(v) Equation of the curves are

and

Slope of the tangent at  =  (say)

Slope of the normal at  is

Point  =

Equation of the tangent is

And Equation of the normal is

### 15. Find the equation of the tangent line to curve  which is:

(a) parallel to the line

(b) perpendicular to the line

Ans. Given:  Equation of the curve ……….(i)

Slope of tangent =  …….(ii)

(a) Slope of the line  is

Slope of tangent parallel to this line is also = 2

From eq. (ii),

From eq. (i),

Therefore, point of contact is (2, 7).

Equation of the tangent at (2, 7) is

(b) Slope of the line  is  =

Slope of the required tangent perpendicular to this line =

From eq. (ii),

From eq. (i),

=

Therefore, point of contact is

Equation of the required tangent is

### 16. Show that the tangents to the curve  at the points where  and  are parallel.

Ans. Given: Equation of the curve

Slope of tangent at  =

At the point  Slope of the tangent =

At the point  Slope of the tangent =

Since, the slopes of the two tangents are equal.

Therefore, tangents at  and  are parallel.

### 17. Find the points on the curve  at which the slope of the tangent is equal to the coordinate of the point.

Ans. Given: Equation of the curve ………(i)

Slope of tangent at

……….(ii)

According to question, Slope of the tangent = coordinate of the point

or

or

From eq. (i), at   The point is (0, 0).

And From eq. (i), at  The point is (3, 27).

Therefore, the required points are (0, 0) and (3, 27).

### 18. For the curve  find all points at which the tangent passes through the origin.

Ans. Given: Equation of the curve  ……….(i)

Slope of the tangent at  passing through origin (0, 0)

Substituting this value of  in eq. (i), we get,

or

or

From eq. (i), at

From eq. (i), at

From eq. (i), at

Therefore, the required points are (0, 0), (1, 2) and

### 19. Find the points on the curve  at which the tangents are parallel to axis.

Ans. Equation of the curve  ……….(i)

[tangent is parallel to axis]

From eq. (i),

Therefore, the required points are (1, 2) and

### 20. Find the equation of the normal at the point  for the curve

Ans. Given: Equation of the curve ……….(i)

Slope of the tangent at the point

=

Slope of the normal at the point  =

Equation of the normal at

### 21. Find the equations of the normal to the curve  which are parallel to the line

Ans. Given: Equation of the curve  ….(i)

Slope of the tangent at

Slope of the normal to the curve at

……….(ii)

But Slope of the normal (given) =

=

From eq. (i), at

at

Therefore, the points of contact are (2, 18) and

Equation of the normal at (2, 18) is

And Equation of the normal at  is

### 22. Find the equation of the tangent and normal to the parabola  at the point

Ans. Given: Equation of the parabola  ……….(i)

Differentiating eq. (i) w.r.t

Slope of the tangent at the point  =

Slope of the normal =

Equation of the tangent at the point

And Equation of the normal at the point

### 23. Prove that the curves  and  cut at right angles if

Ans. Given: Equations of the curves are  …..(i) and ……….(ii)

Substituting the value of  in eq. (ii), we get

Putting the value of  in eq. (i), we get

Therefore, the point of intersection  is = ……….(iii)

Differentiating eq. (i) w.r.t

……….(iv)

Differentiating eq. (ii) w.r.t

……….(v)

According to the question,

[From eq. (iii)]

[Cubing both sides]

### 24. Find the equations of the tangent and normal to the hyperbola x2a2−y2b2=1x2a2−y2b2=1 at the point

Ans. Given: Equation of the hyperbola ……….(i)

……….(ii)

Slope of tangent at  is

Equation of the tangent at  is

……….(iii)

Since  lies on the hyperbola (i), therefore,

From eq. (iii),

Now, Slope of normal at  =

Equation of the normal at  is yy0=a2y0b2x0(xx0)y−y0=−a2y0b2x0(x−x0)

Dividing both sides by

### 25. Find the equation of the tangent to the curve  which is parallel to the line

Ans. Given: Equation of the curve  ……….(i)

Slope of the tangent at point  is

dydx=12(3x2)1/2d(3x2)dxdydx=12(3x−2)−1/2d(3x−2)dx

……….(ii)

Again slope of the line  is ……….(iii)

According to the question,  [Parallel lines have same slope]

From eq. (i),

Therefore, point of contact is .

Equation of the required tangent is

### Choose the correct answer in Exercises 26 and 27.

26. The slope of the normal to the curve  at  is:

(A) 3

(B)

(C)

(D)

Ans. Given: Equation of the curve ……….(i)

Slope of the tangent at point  is

Slope of the tangent at  (say)

Slope of the normal =

Therefore, option (D) if correct.

### 27. The line  is a tangent to the curve  at the point:

(A) (1, 2)

(B) (2, 1)

(C)

(D)

Ans. Given:  Equation of the curve  ……….(i)

Slope of the tangent at point  is 2ydydx=42ydydx=4

……….(ii)

Slope of the line

is ……….(iii)

From eq. (ii) and (iii),

From eq. (i),

Therefore, required point is (1, 2).

Therefore, option (A) is correct.