Chapter 6 Application of Derivatives Ex 6.3 Solutions
Question - 11 : - Find the equation of alllines having slope 2 which are tangents to the curve
Answer - 11 : - The equation of the givencurve is
The slope of the tangent to the given curveat any point (x, y) is given by,
If the slope of the tangent is 2, then we have:
Hence, there is no tangent to the given curve having slope2.
Answer - 12 : - The equation of the givencurve is
The slope of the tangent to the given curveat any point (x, y) is given by,
If the slope of the tangent is 0, then we have:
When x =1, 
∴The equation of the tangent through
is given by,
Hence, the equation of therequired line is
at which the tangents are
(i) parallel to x-axis (ii) parallelto y-axis
Answer - 13 : - The equation of the givencurve is
On differentiating both sides with respectto x, we have:
(i) The tangent is parallelto the x-axis if the slope of the tangent is i.e., 0
which is possible if x =0.
Then,
for x = 0Hence, the points at which the tangents areparallel to the x-axis are
(0, 4) and (0, − 4).
(ii) The tangent is parallel to the y-axisif the slope of the normal is 0, which gives⇒ y = 0.
Then, for y = 0. Hence, the points at which the tangents areparallel to the y-axis are (3, 0) and (− 3, 0).
Question - 14 : - Find the equations of the tangent and normal to the givencurves 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) y = x2 at(0, 0)
(v) x = cos t, y =sin t at 
Answer - 14 : -
(i) The equation of the curve is y = x4 −6x3 + 13x2 − 10x +5.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (0, 5) is −10. Theequation 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 
Therefore, the equation of the normal at (0, 5) is givenas:
(ii) The equation of the curve is y = x4 −6x3 + 13x2 − 10x +5.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (1, 3) is 2. Theequation of the tangent is given as:
The slope of the normal at(1, 3) is
Therefore, the equation of the normal at (1, 3) is givenas:
(iii) The equation of the curve is y = x3.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (1, 1) is 3 and theequation of the tangent is given as:
The slope of the normal at(1, 1) is
Therefore, the equation of the normal at (1, 1) is givenas:
(iv) The equation of the curve is y = x2.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (0, 0) is 0 and theequation of the tangent is given as:
y − 0 =0 (x − 0)
⇒ y = 0
The slope of the normal at (0, 0) is
, which is not defined.
Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by
(v) The equation of the curve is x =cos t, y = sin t.
∴The slope of the tangent at
is −1.
When
Thus, the equation of thetangent to the given curve at 
The slope of the normal atis 
Therefore, the equation ofthe normal to the given curve at
Question - 15 : - Find the equation of the tangent line to thecurve y = x2 − 2x +7 which is
(a) parallel to the line 2x − y +9 = 0
(b) perpendicular to the line 5y −15x = 13.
Answer - 15 : - The equation of the givencurve is
On differentiating with respect to x,we get:
(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.
Therefore, we have:
2 = 2x − 2
Now, x = 2
y = 4 −4 + 7 = 7
Thus, the equation of the tangent passing through (2, 7)is given by,
Hence, the equation of thetangent line to the given curve (which is parallel to line 2x − y +9 = 0) is
.
(b) The equation of the line is 5y −15x = 13.
5y − 15x = 13 ⇒ 
This is of the form y = mx + c.
∴Slope of the line = 3
If a tangent is perpendicular to the line 5y −15x = 13, then the slope of the tangent is
Thus, the equation of thetangent passing through
is given by,
Hence, the equation of thetangent line to the given curve (which is perpendicular to line 5y −15x = 13) is
Show that the tangents to the curve y =7x3 + 11 at the points where x = 2and x = −2 are parallel.
Answer - 16 : -
The equation of the given curve is y =7x3 + 11.
The slope of the tangent toa curve at (x0, y0)is
. Therefore, the slope of the tangent at thepoint where x = 2 is given by,
It is observed that the slopes of thetangents at the points where x = 2 and x = −2are equal.
Hence, the two tangents are parallel.
Question - 17 : - Find the points on the curve y = x3 atwhich the slope of the tangent is equal to the y-coordinate of thepoint.
Answer - 17 : -
The equation of the given curve is y = x3.
The slope of the tangent at the point (x, y)is given by,
When the slope of the tangent is equal tothe 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, then y =0 and when x = 3, then y = 3(3)2 =27.
Hence, the required points are (0, 0) and (3, 27).
Question - 18 : - For the curve y = 4x3 −2x5, find all the points at which the tangents passes throughthe origin.
Answer - 18 : -
The equation of the given curve is y =4x3 − 2x5.
Therefore, the slope of the tangent at apoint (x, y) is 12x2 −10x4.
The equation of the tangent at (x, y)is given by,
When the tangent passes through the origin(0, 0), then X = Y = 0.
Therefore, equation (1) reduces to:
Also, we have
When x =0, y =
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).
Question - 19 : - Find the points on the curve x2 + y2 −2x − 3 = 0 at which the tangents are parallel to the x-axis.
Answer - 19 : -
The equation of the given curve is x2 + y2 −2x − 3 = 0.
On differentiating with respect to x,we have:
Now, the tangents are parallel to the x-axisif the slope of the tangent is 0.
But, x2 + y2 −2x − 3 = 0 for x = 1.
y2 =4⇒
Hence, the points at which the tangents areparallel to the x-axis are (1, 2) and (1, −2).
Question - 20 : - Find the equation of the normal at the point(am2, am3) for the curve ay2 = x3.
Answer - 20 : -
The equation of the given curve is ay2 = x3.
On differentiating with respect to x, wehaveThe slope of a tangent to thecurve at (x0, y0)is Theslope of the tangent to the given curve at (am2, am3) is ∴ Slope of normal at (am2, am3)= 
Hence, the equation of the normal at (am2, am3) isgiven by,
y − am3 =
