Class 12 Maths NCERT Solutions for Chapter 6 Application of Derivatives Exercise 6.3
Application of Derivatives Exercise 6.3 Solutions
1. Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4.
Solution
The given curve is y = 3x4 − 4x.
Then, the slope of the tangent to the given curve at x = 4 is given by, 
= 12(4)3 - 4 = 12(64) - 4 = 764
2. Find the slope of the tangent to the curve y = (x -1)/(x - 2), x != 2 at x = 10.
Solution
3. Find the slope of the tangent to curve y = x3 − x + 1 at the point whose x-coordinate is 2.
Solution
The given curve is y = x3 - x + 1.
∴ dy/dx = 3x2 - 1.
The slope of the tangent to a curve a (x0 , y0) is dy/dx .
It is given that x0 = 2.
Hence, the slope of the tangent at the point where the x - coordinate is 2 is given by,
= 3(2)2 - 1 = 12 - 1 = 11
∴ dy/dx = 3x2 - 1.
The slope of the tangent to a curve a (x0 , y0) is dy/dx .
It is given that x0 = 2.
Hence, the slope of the tangent at the point where the x - coordinate is 2 is given by,
= 3(2)2 - 1 = 12 - 1 = 11
4. Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x-coordinate is 3.
Solution
The given curve is y = x3 − 3x + 2 .
∴ dy/dx = 3x2 - 3
The slope of the tangent to a curve at (x0 , y0) is dy/dx](x0 , y0) .
Hence, the slope of the tangent at the point where the x - coordinate is 3 is given by,
dy/dx]x-3 = 3x2 - 3]x-3 = 3(3)2 - 3 = 2 -3 = 24
∴ dy/dx = 3x2 - 3
The slope of the tangent to a curve at (x0 , y0) is dy/dx](x0 , y0) .
Hence, the slope of the tangent at the point where the x - coordinate is 3 is given by,
dy/dx]x-3 = 3x2 - 3]x-3 = 3(3)2 - 3 = 2 -3 = 24
5. Find the slope of the normal to the curve x = acos3θ, y = asin3θ at θ = π/4 .
Solution
It is given that x = acos3θ, and y = asin3θ .
6. Find the slope of the normal to the curve x = 1 - a sin θ and y = b cos2 θ at θ = π/2.
Solution
It is given that x = 1 - a sin θ and y = b cos2 θ .
7. Find points at which the tangent to the curve y = x3 − 3x2 − 9x + 7 is parallel to the x-axis.
Solution
The equation of the given curve is y = x3 − 3x2 − 9x + 7 .
∴ dy/dx = 3x2 - 6x - 9
Now, the tangent is parallel to the x- axis if the slope of the tangent is zero.
∴ 3x2 - 6x - 9 = 0
⇒ x2 - 2x - 3 = 0
⇒ (x -3)(x + 1) = 0
⇒ x = 3 or x = -1
When x = 3, y = (3)3 - 3(3)2 - 9(3) + 7 = 27 - 27 - 27 + 7 = -20.
When x = -1, y = (-1)3 - 3(-1)2 - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12.
Hence, the points at which the tangent is parallel to the x - axis are (3, -20) and (-1, 12).
∴ dy/dx = 3x2 - 6x - 9
Now, the tangent is parallel to the x- axis if the slope of the tangent is zero.
∴ 3x2 - 6x - 9 = 0
⇒ x2 - 2x - 3 = 0
⇒ (x -3)(x + 1) = 0
⇒ x = 3 or x = -1
When x = 3, y = (3)3 - 3(3)2 - 9(3) + 7 = 27 - 27 - 27 + 7 = -20.
When x = -1, y = (-1)3 - 3(-1)2 - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12.
Hence, the points at which the tangent is parallel to the x - axis are (3, -20) and (-1, 12).
8. Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution
If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.
The slope of the chord is (4- 0)/(4 - 2) = 4/2 = 2.
The slope of the chord is (4- 0)/(4 - 2) = 4/2 = 2.
Now, the slope of the tangent to the given curve at a point (x, y) is given by,
dy/dx = 2(x - 2)
Since the slope of the tangent = slope of the chord, we have :
2(x - 2) = 2
⇒ x - 2 = 1
dy/dx = 2(x - 2)
Since the slope of the tangent = slope of the chord, we have :
2(x - 2) = 2
⇒ x - 2 = 1
⇒ x = 3
When x = 3, y = (3 - 2)2 = 1.
Hence, the required point is (3, 1).
When x = 3, y = (3 - 2)2 = 1.
Hence, the required point is (3, 1).
9. Find the point on the curve y = x3 − 11x + 5 at which the tangent is y = x − 11.
Solution
The equation of the given curve is y = x3 − 11x + 5 .
The equation of the tangent to the given curve is given as y = x - 11 (which is of the form y = mx + c).
∴ Slope of the tangent = 1
Now, the slope of the tangent to the given curve at the point (x, y) is given by, dy/dx = 3x2 - 11
Then, we have :
3x2 - 11 = 1
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
When x = 2, y = (2)3 - 11(2) + 5 = 8 - 22 + 5 = -9.
When x = -2, y = (-2)3 - 11(-2) +5 = -8 + 22 + 5 = 19.
The equation of the tangent to the given curve is given as y = x - 11 (which is of the form y = mx + c).
∴ Slope of the tangent = 1
Now, the slope of the tangent to the given curve at the point (x, y) is given by, dy/dx = 3x2 - 11
Then, we have :
3x2 - 11 = 1
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
When x = 2, y = (2)3 - 11(2) + 5 = 8 - 22 + 5 = -9.
When x = -2, y = (-2)3 - 11(-2) +5 = -8 + 22 + 5 = 19.
Hence, the required points are (2, -9) and (-2, 19).
10. Find the equation of all lines having slope −1 that are tangents to the curve y = 1/(x-1), x ≠ 1 .
Solution
The equation of the given curve is y = 1/(x-1), x ≠ 1 .
The slope of the tangents to the given curve at any point (x, y) is given by,
dy/dx = -1/(x - 1)2
If the slope of the tangent is -1, then we have :
-1/(x - 1)2 = -1
⇒ (x - 1)2 = 1
⇒ x - 1 = ± 1
⇒ x = 2, 0
When x = 0, y = -1 and when x = 2, y = 1.
Thus, there are two tangents to the given curve having slope -1. These are passing through the points (0, -1) and (2, 1).
∴ The equation of the tangent through (0, -1) is given by,
y - (-1) = -1(x - 0)
⇒ y + 1 = -x
⇒ y + x + 1 = 0
∴ The equation of the tangent through (2, 1) is given by,
y - 1 = -1(x - 2)
⇒ y - 1 = -x + 2
⇒ y + x - 3 = 0
Hence, the equations of the required lines are y + x + 1 = 0 and y + x - 3 = 0.
The slope of the tangents to the given curve at any point (x, y) is given by,
dy/dx = -1/(x - 1)2
If the slope of the tangent is -1, then we have :
-1/(x - 1)2 = -1
⇒ (x - 1)2 = 1
⇒ x - 1 = ± 1
⇒ x = 2, 0
When x = 0, y = -1 and when x = 2, y = 1.
Thus, there are two tangents to the given curve having slope -1. These are passing through the points (0, -1) and (2, 1).
∴ The equation of the tangent through (0, -1) is given by,
y - (-1) = -1(x - 0)
⇒ y + 1 = -x
⇒ y + x + 1 = 0
∴ The equation of the tangent through (2, 1) is given by,
y - 1 = -1(x - 2)
⇒ y - 1 = -x + 2
⇒ y + x - 3 = 0
Hence, the equations of the required lines are y + x + 1 = 0 and y + x - 3 = 0.
11. Find the equation of all lines having slope 2 which are tangents to the curve y = 1/(x - 3), x ≠ 3.
Solution
The equation of the given curve is y = 1/(x - 3), x ≠ 3.
The slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = -1/(x - 3)2
If the slope of the tangent is 2, then we have :
-1/(x- 3)2 = 2
⇒ 2(x - 3)2 = -1
⇒ (x - 3)2 = -1/2
This is not possible since the L.H.S. is positive while the R.H.S. is negative .
Hence, there is no tangent to the given curve having slope 2.
The slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = -1/(x - 3)2
If the slope of the tangent is 2, then we have :
-1/(x- 3)2 = 2
⇒ 2(x - 3)2 = -1
⇒ (x - 3)2 = -1/2
This is not possible since the L.H.S. is positive while the R.H.S. is negative .
Hence, there is no tangent to the given curve having slope 2.
12. Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x2 - 2x + 3) .
Solution
The slope of the tangent to the given curve at any point (x, y) is given by,
⇒ - 2(x - 1) = 0
⇒ x = 1
When x = 1, y = 1/(1 - 2 + 3) = 1/2.
∴ The equation of the tangent through (1, 1/2) is given by,
y - 1/2 = 0 (x - 1)
⇒ y - 1/2 = 0
⇒ y = 1/2
Hence, the equation of the required line is y = 1/2.
⇒ - 2(x - 1) = 0
⇒ x = 1
When x = 1, y = 1/(1 - 2 + 3) = 1/2.
∴ The equation of the tangent through (1, 1/2) is given by,
y - 1/2 = 0 (x - 1)
⇒ y - 1/2 = 0
⇒ y = 1/2
Hence, the equation of the required line is y = 1/2.
13. Find points on the curve x2/9 + y2/16 =1 at which the tangent are
(i) parallel to x - axis
(i) parallel to x - axis
(ii) parallel to y - axis
Solution
The equation of the given curve is x2/9 + y2/16 =1 .
On differentiating both sides with respect to x, we have :
2x/9 + 2y/16 .dy/dx = 0
⇒ dy/dx = -16x/9y
On differentiating both sides with respect to x, we have :
2x/9 + 2y/16 .dy/dx = 0
⇒ dy/dx = -16x/9y
(i) The tangent is parallel to the x - axis if the slope of the tangent is 0 i.e., -16x/9y = 0, which is possible if x = 0.
Then, x2/9 + y2/16 =1 for x = 0
⇒ y2 = 16
Then, x2/9 + y2/16 =1 for x = 0
⇒ y2 = 16
⇒ y = ± 4
Hence, the points at which the tangents are parallel to the x - axis are (0, 4) and (0, -4).
Hence, the points at which the tangents are parallel to the x - axis are (0, 4) and (0, -4).
(ii) The tangent is parallel to the y - axis if the slope of the normal is 0, which gives
Then, x2 /9 + y2 /16 = 1 for y = 0.
⇒ x = ± 3
Hence, the points at which the tangents are parallel to the y - axis are (3, 0) and (-3, 0).
14. Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)
(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) x2 at (0, 0)
(v) x = cos t, y = sin t at t = π/4
(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)
(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) x2 at (0, 0)
(v) x = cos t, y = sin t at t = π/4
Solution
(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5
On differentiating with respect to x, we get :
dy/dx = 4x3 - 18x2 + 26x - 10
dy/dx](0, 5) = -10
Thus, the slope of the tangent at (0, 5) is -10. The equation of the tangent is given as :
y - 5 = -10(x - 0)
⇒ y - 5 = -10x
⇒ 10x + y = 5
The slope of the normal at (0, 5) is -1/[Slope of the tangent at (0, 5)] = 1/10 .
Therefore, the equation of the normal at (0, 5) is given as :
y - 5 = 1/10 (x - 0)
⇒ 10y - 50 = x
⇒ x - 10y + 50 = 0
On differentiating with respect to x, we get :
dy/dx = 4x3 - 18x2 + 26x - 10
dy/dx](0, 5) = -10
Thus, the slope of the tangent at (0, 5) is -10. The equation of the tangent is given as :
y - 5 = -10(x - 0)
⇒ y - 5 = -10x
⇒ 10x + y = 5
The slope of the normal at (0, 5) is -1/[Slope of the tangent at (0, 5)] = 1/10 .
Therefore, the equation of the normal at (0, 5) is given as :
y - 5 = 1/10 (x - 0)
⇒ 10y - 50 = x
⇒ x - 10y + 50 = 0
(ii) The equation of the curve is y = x4 - 6x3 + 13x2 - 10x + 5.
On differentiating with respect to x, we get :
dy/dx = 4x3 - 18x2 + 26x - 10
dy/dx](1, 3) = 4 - 18 + 26 - 10 = 2
Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as :
y - 3 = 2(x - 1)
⇒ y - 3 = 2x - 2
⇒ y = 2x + 1
The slope of the normal at (1, 3) is -1/[Slope of the tangent at (1, 3)] = -1/2 .
dy/dx = 4x3 - 18x2 + 26x - 10
dy/dx](1, 3) = 4 - 18 + 26 - 10 = 2
Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as :
y - 3 = 2(x - 1)
⇒ y - 3 = 2x - 2
⇒ y = 2x + 1
The slope of the normal at (1, 3) is -1/[Slope of the tangent at (1, 3)] = -1/2 .
Therefore, the equation of the normal at (1, 3) is given as :
y - 3 = -1/2 (x - 1)
⇒ 2y - 6 = -x + 1
⇒ x + 2y - 7 = 0
y - 3 = -1/2 (x - 1)
⇒ 2y - 6 = -x + 1
⇒ x + 2y - 7 = 0
(iii) The equation of the curve is y = x3 .
On differentiating with respect to x, we get :
dy/dx = 3x2
On differentiating with respect to x, we get :
dy/dx = 3x2
dy/dx](1, 1) = 3(1)2 = 3
Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as :
y - 1 = 3(x - 1)
⇒ y = 3x - 2
The slope of the normal at (1, 1) is -1/[Slope of the tangent at (1, 1)] = -1/3 .
Therefore, the equation of the normal at (1, 1) is given as :
y - 1 = -1/3 (x - 1)
⇒ 3y - 3 = - x + 1
⇒ x + 3y - 4 = 0
Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as :
y - 1 = 3(x - 1)
⇒ y = 3x - 2
The slope of the normal at (1, 1) is -1/[Slope of the tangent at (1, 1)] = -1/3 .
Therefore, the equation of the normal at (1, 1) is given as :
y - 1 = -1/3 (x - 1)
⇒ 3y - 3 = - x + 1
⇒ x + 3y - 4 = 0
(iv) The equation of the curve is y = x2 .
On differentiating with respect to x, we get :
dy/dx = 2x
dy/dx](0, 0) = 0
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as :
y - 0 = 0(x - 0)
⇒ y = 0
The slope of the normal at (0, 0) is -1/[Slope of the tangent at (0, 0)] = -1/0 , which is not defined.
On differentiating with respect to x, we get :
dy/dx = 2x
dy/dx](0, 0) = 0
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as :
y - 0 = 0(x - 0)
⇒ y = 0
The slope of the normal at (0, 0) is -1/[Slope of the tangent at (0, 0)] = -1/0 , which is not defined.
Therefore, the equation of the normal at (x0 , y0) = (0, 0) is given by
x = x0 = 0.
x = x0 = 0.
(v) The equation of the curve is x = cos t, y = sin t.
x = cos t and y = sin t
∴ dx/dt = - sin t, dy/dt = cos t
x = cos t and y = sin t
∴ dx/dt = - sin t, dy/dt = cos t
∴ The slope of the tangent at t = Ï€/4 is -1.
When t = Ï€/4 , x = 1/√2 and y = 1/√2 .
Thus, the equation of the tangent to the given curve at t = Ï€/4 i.e., at [(1/√2 , 1/√2)] is
y - 1/√2 = -1 (x - 1/√2) .
⇒ x + y - 1/√2 - 1/√2 = 0
⇒ x + y - √2 = 0
The slope of the normal at t = π/4 is -1/[Slope of the tangent at t = π/4] = 1.
Therefore, the equation of the normal to the given curve at t = Ï€/4 i.e., at [(1/√2 , 1/√2)] is
y - 1/√2 = 1 (x - 1/√2)
⇒ x = y
15. Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is
(a) parallel to the line 2x − y + 9 = 0.
(b) perpendicular to the line 5y - 15x = 13
(a) parallel to the line 2x − y + 9 = 0.
(b) perpendicular to the line 5y - 15x = 13
Solution
The equation of the given curve is y = x2 − 2x + 7 .
On differentiating with respect to x, we get :
dy/dx = 2x - 2
On differentiating with respect to x, we get :
dy/dx = 2x - 2
(a) The equation of the line is 2x - y + 9 = 0.
2x - y + 9 = 0 ⇒ y = 2x + 9
This is of the form y = mx + c .
∴ Slope of the line = 2
If a tangent is parallel to the line 2x - y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
2x - y + 9 = 0 ⇒ y = 2x + 9
This is of the form y = mx + c .
∴ Slope of the line = 2
If a tangent is parallel to the line 2x - y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have :
2 = 2x - 2
⇒ 2x = 4
⇒ x = 2
Now, x = 2
⇒ y = 4 - 4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7) is given by,
y - 7 = 2(x - 2)
⇒ y- 2x - 3 = 0
Hence, the equation of the tangent line to the given curve (which is parallel to line 2x - y + 9 = 0 ) is y - 2x - 3 = 0.
2 = 2x - 2
⇒ 2x = 4
⇒ x = 2
Now, x = 2
⇒ y = 4 - 4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7) is given by,
y - 7 = 2(x - 2)
⇒ y- 2x - 3 = 0
Hence, the equation of the tangent line to the given curve (which is parallel to line 2x - y + 9 = 0 ) is y - 2x - 3 = 0.
(b) The equation of the line is 5y - 15x = 13.
5y - 15x = 13 ⇒ y = 3x + 13/5
This is of the form y = 2x + c .
∴ Slope of the line = 3
If a tangent is perpendicular to the line 5y - 15x = 13, then the slope of the tangent is -1/[slope of the line] = -1/3 .
5y - 15x = 13 ⇒ y = 3x + 13/5
This is of the form y = 2x + c .
∴ Slope of the line = 3
If a tangent is perpendicular to the line 5y - 15x = 13, then the slope of the tangent is -1/[slope of the line] = -1/3 .
⇒ 2x - 2 = -1/3
⇒ 2x = -1/3 + 2
⇒ 2x = 5/3
⇒ x = 5/6
Now, x = 5/6
⇒ y = (25/36) -(10/6) + 7 = (25 - 60 + 252)/36 = 217/36
Thus, the equation of the tangent passing through (5/6, 217/36) is given by,
⇒ 36y - 217 = -2(6x - 5)
⇒ 36y - 217 = -12x + 10
⇒ 36y + 12x - 227 = 0
Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y - 15x = 13) is 36y + 12x - 227 = 0.
⇒ 2x = -1/3 + 2
⇒ 2x = 5/3
⇒ x = 5/6
Now, x = 5/6
⇒ y = (25/36) -(10/6) + 7 = (25 - 60 + 252)/36 = 217/36
Thus, the equation of the tangent passing through (5/6, 217/36) is given by,
⇒ 36y - 217 = -2(6x - 5)
⇒ 36y - 217 = -12x + 10
⇒ 36y + 12x - 227 = 0
Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y - 15x = 13) is 36y + 12x - 227 = 0.
16. Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.
Solution
The equation of the given curve is y = 7x3 + 11.
∴ dy/dx = 21x2
The slope of the tangent to a curve at (x0 , y0) is dy/dx](x0 ,y0)
Therefore, the slope of the tangent at the point where x = 2 is given by,
dy/dx]x = 2 = 21(2)2 = 84
And, the slope of the tangent at the point where x = -2 is given by,
dy/dx]x = -2 = 21(2)2 = 84
It is observed that the slopes of the tangents at the points where x = 2 and x = -2 are equal.
∴ dy/dx = 21x2
The slope of the tangent to a curve at (x0 , y0) is dy/dx](x0 ,y0)
Therefore, the slope of the tangent at the point where x = 2 is given by,
dy/dx]x = 2 = 21(2)2 = 84
And, the slope of the tangent at the point where x = -2 is given by,
dy/dx]x = -2 = 21(2)2 = 84
It is observed that the slopes of the tangents at the points where x = 2 and x = -2 are equal.
Hence, the two tangents are parallel.
17. Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.
Solution
The equation of the given curve is y = x3.
∴ dy/dx = 3x2
The slope of the tangent at the point (x, y) is given by,
dy/dx](x, y) = 3x2
When the slope of the tangent is equal to the y - coordinate of the point, then y = 3x2.
Also, we have y = x3 .
∴ 3x2 = x3
⇒ x2 (x - 3) = 0
⇒ x = 0, x = 3
When x = 0, they y = 0 and when x = 3, then y = 3 (3)2 = 27.
Hence, the required points are (0, 0) and (3, 27).
∴ dy/dx = 3x2
The slope of the tangent at the point (x, y) is given by,
dy/dx](x, y) = 3x2
When the slope of the tangent is equal to the y - coordinate of the point, then y = 3x2.
Also, we have y = x3 .
∴ 3x2 = x3
⇒ x2 (x - 3) = 0
⇒ x = 0, x = 3
When x = 0, they y = 0 and when x = 3, then y = 3 (3)2 = 27.
Hence, the required points are (0, 0) and (3, 27).
18. For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin.
Solution
The equation of the given curve is y = 4x3 − 2x5 .
∴ dy/dx = 12x2 - 10x4
Therefore, the slope of the tangent at a point (x, y) is 12x2 - 10x4 .
The equation of the tangent at (x, y) is given by,
Y - y = (12x2 - 10x4 )(X - x) ...(1)
When the tangent passes through the origin (0, 0), then X = Y = 0.
Therefore, equation (1) reduces to :
-y = (12x2 - 10x4 )(-x)
y = 12x3 - 10x5
Also, we have y = 4x3 - 2x5 ,
∴ 12x3 - 10x5 = 4x3 - 2x5
⇒ 8x5 - 8x3 = 0
⇒ x5 - x3 = 0
⇒ x3 (x2 - 1) = 0
⇒ x = 0, ± 1
When x = 0, y = 4(0)3 - 2(0)5 = 0.
When x = 1, y = 4(1)3 - 2(1)5 = 2.
When x = -1, y = 4(-1)3 - 2(-1)5 = -2.
Hence, the required points are (0, 0), (1, 2) and (-1, -2).
∴ dy/dx = 12x2 - 10x4
Therefore, the slope of the tangent at a point (x, y) is 12x2 - 10x4 .
The equation of the tangent at (x, y) is given by,
Y - y = (12x2 - 10x4 )(X - x) ...(1)
When the tangent passes through the origin (0, 0), then X = Y = 0.
Therefore, equation (1) reduces to :
-y = (12x2 - 10x4 )(-x)
y = 12x3 - 10x5
Also, we have y = 4x3 - 2x5 ,
∴ 12x3 - 10x5 = 4x3 - 2x5
⇒ 8x5 - 8x3 = 0
⇒ x5 - x3 = 0
⇒ x3 (x2 - 1) = 0
⇒ x = 0, ± 1
When x = 0, y = 4(0)3 - 2(0)5 = 0.
When x = 1, y = 4(1)3 - 2(1)5 = 2.
When x = -1, y = 4(-1)3 - 2(-1)5 = -2.
Hence, the required points are (0, 0), (1, 2) and (-1, -2).
19. Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.
Solution
The equation of the given curve is x2 + y2 − 2x − 3 = 0 .
On differentiating with respect to x, we have :
2x + 2y(dy/dx) - 2 = 0
⇒ y. (dy/dx) = 1 - x
⇒ dy/dx = (1 - x)/y
Now, the tangents are parallel to the x - axis if the slope of the tangent is 0.
2x + 2y(dy/dx) - 2 = 0
⇒ y. (dy/dx) = 1 - x
⇒ dy/dx = (1 - x)/y
Now, the tangents are parallel to the x - axis if the slope of the tangent is 0.
∴ (1 - x)/y = 0
⇒ 1 - x = 0
⇒ x = 1
But, x2 + y2 -2x - 3 = 0 for x = 1.
⇒ y2 = 4
But, x2 + y2 -2x - 3 = 0 for x = 1.
⇒ y2 = 4
⇒ y = ± 2
Hence, the points at which the tangents are parallel to the x - axis are (1, 2) and (1, -2).
Hence, the points at which the tangents are parallel to the x - axis are (1, 2) and (1, -2).
20. Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.
Solution
The equation of the given curve is ay2 = x3 .
On differentiating with respect to x, we have :
On differentiating with respect to x, we have :
2ay . (dy/dx) = 3x2
⇒ dy/dx = 3x2 /2ay
The slope of a tangent to the curve at (x0, y0) is
⇒ The slope of the tangent to the given curve at (am2, am3) is
∴ Slope of normal at (am2, am3)
= -1/[slope of the tangent at (am2, am3) = -2/3m
Hence, the equation of the normal at (am2, am3) is given by,
y - am3 = -2/3m (x - am2)
⇒ 3my - 3am4 = -2x + 2am2
⇒ 2x + 3my - am2 (2 + 3m2) = 0
⇒ dy/dx = 3x2 /2ay
The slope of a tangent to the curve at (x0, y0) is
⇒ The slope of the tangent to the given curve at (am2, am3) is
∴ Slope of normal at (am2, am3)
= -1/[slope of the tangent at (am2, am3) = -2/3m
Hence, the equation of the normal at (am2, am3) is given by,
y - am3 = -2/3m (x - am2)
⇒ 3my - 3am4 = -2x + 2am2
⇒ 2x + 3my - am2 (2 + 3m2) = 0
21. Find the equation of the normal to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution
The equation of the given curve is y = x3 + 2x + 6.
The slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = 3x2 + 2
∴ Slope of the normal to the given curve at any point (x, y)
= -1/[Slope of the tangent at the point (x, y)]
= -1/(3x2 + 2)
The equation of the given line is x + 14y + 4 = 0.
x + 14y + 4 ⇒ y = -(1/14)x - 4/14 (which is of the form y = mx + c)
∴ Slope of the given line = -1/14
If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.
The slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = 3x2 + 2
∴ Slope of the normal to the given curve at any point (x, y)
= -1/[Slope of the tangent at the point (x, y)]
= -1/(3x2 + 2)
The equation of the given line is x + 14y + 4 = 0.
x + 14y + 4 ⇒ y = -(1/14)x - 4/14 (which is of the form y = mx + c)
∴ Slope of the given line = -1/14
If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.
∴ -1/(3x2 +2) = -1/14
⇒ 3x3 + 2 = 14
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
⇒ 3x3 + 2 = 14
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
When x = 2, y = 8 + 4 + 6 = 18.
When x = -2, y = -8 - 4 + 6 = -6.
Therefore, there are two normals to the given curve with slope -1/14 and passing through the points (2, 18) and (-2, -6).
thus, the equation of the normal through (2, 18) is given by,
y - 18 = (-1/14)(x - 2)
⇒ 14y - 252 = -x + 2
⇒ x + 14y - 254 = 0
And, the equation of the normal through (-2, -6) is given by,
y -(-6) = -1/14 (x - 2)
⇒ 14y - 252 = -x + 2
⇒ x + 14y - 254 = 0
And, the equation of the normal through (-2, -6) is given by,
y - (-6) = (-1/14) [x-(-2)]
⇒ y + 6 = (-1/14) (x + 2)
⇒ 14y + 84 = -x - 2
⇒ x + 14y + 86 = 0
Hence, the equations of the normals to the given curve (which are parallel to the given line ) are x + 14y - 254 = 0 and x + 14y + 86 = 0.
When x = -2, y = -8 - 4 + 6 = -6.
Therefore, there are two normals to the given curve with slope -1/14 and passing through the points (2, 18) and (-2, -6).
thus, the equation of the normal through (2, 18) is given by,
y - 18 = (-1/14)(x - 2)
⇒ 14y - 252 = -x + 2
⇒ x + 14y - 254 = 0
And, the equation of the normal through (-2, -6) is given by,
y -(-6) = -1/14 (x - 2)
⇒ 14y - 252 = -x + 2
⇒ x + 14y - 254 = 0
And, the equation of the normal through (-2, -6) is given by,
y - (-6) = (-1/14) [x-(-2)]
⇒ y + 6 = (-1/14) (x + 2)
⇒ 14y + 84 = -x - 2
⇒ x + 14y + 86 = 0
Hence, the equations of the normals to the given curve (which are parallel to the given line ) are x + 14y - 254 = 0 and x + 14y + 86 = 0.
22. Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at).
Solution
The equation of the given parabola is y2 = 4ax.
On differentiating y2 = 4ax with respect to x, we have:
2y . (dy/dx) = 4a
⇒ dy/dx = 2a/y
∴ The slope of the tangent at (at2 , 2at) is
Then, the equation of the tangent at (at2 , 2at) is given by,
y - 2at = 1/t (x - at2 )
⇒ ty - 2at2 = x - at2
⇒ ty = x + at2
Now, the slope of the normal at (at2 , 2at) is given by,
-1/[Slope of the tangent at (at2 , 2at)] = -t
Thus, the equation of the normal at (at2 , 2at) is given as :
y - 2at = -t(x - at2 )
⇒ y - 2at = -tx + at3
⇒ y = -tx + 2at + at3
On differentiating y2 = 4ax with respect to x, we have:
2y . (dy/dx) = 4a
⇒ dy/dx = 2a/y
∴ The slope of the tangent at (at2 , 2at) is
Then, the equation of the tangent at (at2 , 2at) is given by,
y - 2at = 1/t (x - at2 )
⇒ ty - 2at2 = x - at2
⇒ ty = x + at2
Now, the slope of the normal at (at2 , 2at) is given by,
-1/[Slope of the tangent at (at2 , 2at)] = -t
Thus, the equation of the normal at (at2 , 2at) is given as :
y - 2at = -t(x - at2 )
⇒ y - 2at = -tx + at3
⇒ y = -tx + 2at + at3
23. Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]
Solution
The equations of the given curves are given as x = y2 and xy = k.
Putting x = y2 in xy = k, we get :
y3 = k
y3 = k
⇒ y = k1/3
∴ x = k2/3
Thus, the point of intersection of the given curves is (k2/3 , k1/3 ).
Differentiating x = y2 with respect to x, we have :
1 = 2y. (dy/dx)
∴ x = k2/3
Thus, the point of intersection of the given curves is (k2/3 , k1/3 ).
Differentiating x = y2 with respect to x, we have :
1 = 2y. (dy/dx)
⇒ dy/dx = 1/2y
Therefore, the slope of the tangent to the curve x = y2 at
On differentiating xy = k with respect to x, we have:
Therefore, the slope of the tangent to the curve x = y2 at
On differentiating xy = k with respect to x, we have:
x. (dy/dx) + y = 0
⇒ dy/dx = -y/x
∴ Slope of the tangent to the curve xy = k at (k2/3, k1/3) is given by,
We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at (k2/3 , k1/3 ) are perpendicular to each other.
∴ Slope of the tangent to the curve xy = k at (k2/3, k1/3) is given by,
We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at (k2/3 , k1/3 ) are perpendicular to each other.
This implies that we should have the product of the tangents as -1.
Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at (k2/3 , k1/3) is -1.
Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at (k2/3 , k1/3) is -1.
Hence, the given two curves cut at right angels if 8k2 = 1.
24. Find the equations of the tangent and normal to the hyperbola x2/a2 - y2/b2 at the point (x0, y0).
Solution
Differentiating x2 /a2 - y2 /b2 = 1 with respect to x, we have :
25. Find the equation of the tangent to the curve y = √(3x-2) which is parallel to the line 4x − 2y + 5 = 0.
Solution
The equation of the given curve is y = √(3x-2) .
The slope of the tangent to the given curve at any point (x, y) is given by,
The slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = 3/2√(3x - 2)
The equation of the given line is 4x - 2y + 5 = 0.
4x - 2y + 5 = 0 ⇒ y = 2x + 5/2 which is of the form y = mx + c)
∴ Slope of the line = 2
Now, the tangent to the given curve is parallel to the line 4x -2y - 5 = 0 if the slope of the tangent is equal to the slope of the line.
The equation of the given line is 4x - 2y + 5 = 0.
4x - 2y + 5 = 0 ⇒ y = 2x + 5/2 which is of the form y = mx + c)
∴ Slope of the line = 2
Now, the tangent to the given curve is parallel to the line 4x -2y - 5 = 0 if the slope of the tangent is equal to the slope of the line.
3/[2√(3x - 2)] = 2
⇒ √(3x - 2) = 3/4
⇒ 3x - 2 = 9/16
⇒ 3x = 9/16 + 2 = 41/16
⇒ x = 41/48
⇒ 24y - 18 = 48x - 41
⇒ 48x - 24y = 23
Hence, the equation of the required tangent is 48x - 24y = 23
⇒ √(3x - 2) = 3/4
⇒ 3x - 2 = 9/16
⇒ 3x = 9/16 + 2 = 41/16
⇒ x = 41/48
⇒ 24y - 18 = 48x - 41
⇒ 48x - 24y = 23
Hence, the equation of the required tangent is 48x - 24y = 23
26. The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) −3
(D) -1/3
(B) 1/3
(C) −3
(D) -1/3
Solution
The equation of the given curve is y = 2x2 + 3sin x.
Slope of the tangent to the given curve at x = 0 is given by,
dy/dx]x = 0 = 4x + 3 cos x]x = 0 = 0 + 3 cos 0 = 3
Hence, the slope of the normal to the given curve at x = 0 is
-1/[Slope of the tangent at x = 0] = -1/3 .
The correct answer is D.
dy/dx]x = 0 = 4x + 3 cos x]x = 0 = 0 + 3 cos 0 = 3
Hence, the slope of the normal to the given curve at x = 0 is
-1/[Slope of the tangent at x = 0] = -1/3 .
The correct answer is D.
27. The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 2)
(B) (2, 1)
(C) (1, −2)
(D) (−1, 2)
(A) (1, 2)
(B) (2, 1)
(C) (1, −2)
(D) (−1, 2)
Solution
The equation of the given curve is y2 = 4x .
Differentiating with respect to x, we have :
Differentiating with respect to x, we have :
2y. dy/dx = 4 ⇒ dy/dx = 2/y
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = 2/y
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx =2/y
The given line is y = x + 1 (which is of the form y = mx + c)
∴ Slope of the line = 1
The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx = 2/y
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by,
dy/dx =2/y
The given line is y = x + 1 (which is of the form y = mx + c)
∴ Slope of the line = 1
The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.
Thus, we must have :
2/y = 1
⇒ y = 2
Now,
⇒ y = 2
Now,
y = x + 1
⇒ x = y- 1
⇒ x = 2 - 1 = 1
Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).
The correct answer is A.
Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).
The correct answer is A.