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 
……….(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
at 
……….(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.
……….(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.
……….(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
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
at 
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.
……….(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).
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
……….(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
……….(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
= 
= 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
……….(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
……….(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 
slope of tangent = at (0, 5)
=
(say)
Slope of the normal at (0, 5) is 
Equation of the tangent at (0, 5) is y−5=−10(x−0)y−5=−10(x−0)
y−5=−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 
……….(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.
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.
………(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.
find all points at which the tangent passes through the origin.
……….(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.
……….(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
……….(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
….(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
……….(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
…..(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
……….(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 y−y0=−a2y0b2x0(x−x0)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
……….(i) Slope of the tangent at point is 
dydx=12(3x−2)−1/2d(3x−2)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) 
……….(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:
at the point:
(A) (1, 2)
(B) (2, 1)
(C) 
(D) 
……….(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.