NCERT Solutions class 12 Maths Exercise 6.3

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NCERT Solutions class 12 Maths Exercise 6.3 Class 12 Maths Class book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

Download NCERT solutions for Applications of Derivatives as PDF.

NCERT Solutions class 12 Maths Exercise 6.3

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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