Chapter 6 Application of Derivatives Ex 6.3 Solutions
Question - 21 : - Find the equation of the normals to thecurve y = x3 + 2x +6 which are parallel to the line x + 14y + 4 = 0.
Answer - 21 : -
The equation of the given curve is y = x3 +2x + 6.
The slope of the tangent to the given curveat any point (x, y) is given by,
∴ Slope of the normal to the given curve at any point (x, y)=
The equation of the given line is x +14y + 4 = 0.
x +14y + 4 = 0 ⇒ 
(which is of the form y = mx + c)
∴Slope of the given line = 
If the normal is parallel to the line, then we must havethe slope of the normal being equal to the slope of the line.
When x = 2, y =8 + 4 + 6 = 18.
When x = −2, y =− 8 − 4 + 6 = −6.
Therefore, there are two normals to the given curvewith slopeand passing through the points (2, 18)
and (−2, −6).
Thus, the equation of the normal through (2, 18) isgiven by,And, the equation of the normal through (−2, −6) is givenby,
Hence, the equations of thenormals to the given curve (which are parallel to the given line) are
Find the equations of the tangent and normalto the parabola y2 = 4ax atthe point (at2, 2at).
Answer - 22 : -
The equation of the given parabola is y2 =4ax.
On differentiating y2 =4ax with respect to x, we have:
∴The slope of the tangent at
is 
Then, the equation of thetangent atis given by,
y − 2at =
Now, the slope of thenormal atis given by,
Thus, the equation of the normal at (at2, 2at)is given as:
Question - 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 thecurves at the point of intersection are perpendicular to each other.]
Answer - 23 : - The equations of the givencurves are given as
Putting x = y2 in xy = k,we get:
Thus, the point ofintersection of the given curves is
Differentiating x = y2 withrespect to x, we have:
Therefore, the slope of thetangent to the curve x = y2 atis 
On differentiating xy = k withrespect to x, we have:
∴ Slope of the tangent to the curve xy = k atis given by,
We know that two curvesintersect at right angles if the tangents to the curves at the point ofintersection i.e., at
are perpendicular to each other.
This implies that we should have the product of thetangents as − 1.
Thus, the given two curves cut at right angles if theproduct of the slopes of their respective tangents at is −1.
Hence, the given two curves cut at rightangels if 8k2 = 1.
Question - 24 : - Find the equations of thetangent and normal to the hyperbola 
at the point
Answer - 24 : - Differentiating
with respect to x, we have:
Therefore, the slope of thetangent atis 
Then, the equation of thetangent atis given by, Now, the slope of thenormal atis given by, Hence, the equation of thenormal atis given by,
Question - 25 : - Find the equation of thetangent to the curve 
which is parallel to the line 4x −2y + 5 = 0.
Answer - 25 : - The equation of the givencurve is
The slope of the tangent to the given curveat any point (x, y) is given by,
The equation of the given line is 4x −2y + 5 = 0.
4x − 2y + 5 = 0 ⇒ 
(which is of the form
∴Slope of the line = 2
Now, the tangent to the given curve isparallel to the line 4x − 2y − 5 = 0 if the slope ofthe tangent is equal to the slope of the line.
∴Equation of the tangent passing through thepoint 
is given by,
Hence, the equation of therequired tangent is
The slope of the normal to the curve y =2x2 + 3 sin x at x = 0is
(A) 3 (B) 
(C) −3 (D) 
Answer - 26 : - The equation of the givencurve is
Slope of the tangent to the given curveat x = 0 is given by,
Hence, the slope of the normal to the givencurve at x = 0 is
The correct answer is D.
The line y = x +1 is a tangent to the curve y2 = 4x atthe point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)
Answer - 27 : - The equation of the givencurve is
Differentiating with respect to x,we have:
Therefore, the slope of the tangent to thegiven curve at any point (x, y) is given by,
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 theslope of the tangent. Also, the line must intersect the curve.
Thus, we must have:
Hence, the line y = x +1 is a tangent to the given curve at the point (1, 2).
The correct answer is A.