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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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