**Download NCERT solutions for Applications of Derivatives as PDF.**

**NCERT Solutions for Class 12 Maths Application of Derivatives**** **

**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

### NCERT Solutions class 12 Maths Exercise 6.3

**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)

=

=

### NCERT Solutions class 12 Maths Exercise 6.3

**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

### NCERT Solutions class 12 Maths Exercise 6.3

**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

### NCERT Solutions class 12 Maths Exercise 6.3

**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

### NCERT Solutions class 12 Maths Exercise 6.3

**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

=

=

### NCERT Solutions class 12 Maths Exercise 6.3

**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

### NCERT Solutions class 12 Maths Exercise 6.3

**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 =

From eq. (ii),

From eq. (i),

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

**14. Find the equation of the tangents 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)**

**Ans. (i)** Equation of the curve

Now value of at (0, 5)

At (say)

Slope of the normal at (0, 5) is

Equation of the tangent at (0, 5) is

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

**(ii)** Equation of the curve

Now value of at (1, 3)

At (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)

Now value of at (1, 1)

At = (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)

Now value of at (0, 0)

At = (say)

Equation of the tangent at (0, 0) is

And normal at (0, 0) is axis.

**(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 point 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)

Slope of the tangent at

=

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 equation of the tangent and normal to the hyperbola 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

Dividing both sides by

### NCERT Solutions class 12 Maths Exercise 6.3

**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

= ……….(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

……….(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.

## NCERT Solutions class 12 Maths Exercise 6.3

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