RD Chapter 16 Tangents and Normals Ex 16.1 Solutions
Question - 1 : - Find the Slopes of the tangent and the normal to the following curves at the indicated points:(ii) y = √x at x =9
(iii) y = x3 –x at x = 2
(iv) y = 2x2 +3 sin x at x = 0
(v) x = a (θ –sin θ), y = a (1 + cos θ) at θ = -π /2
(vi) x = a cos3 θ,y = a sin3 θ at θ = π /4
(vii) x = a(θ – sin θ), y = a (1 – cos θ) at θ = π /2
(viii) y = (sin 2x+ cot x + 2)2 at x = π /2
(ix) x2 +3y + y2 = 5 at (1, 1)
(x) x y = 6 at (1,6)
Answer - 1 : - (i)
(ii)
⇒ TheSlope of the normal = – 6
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
Answer - 2 : -
⇒ –a – 1 = 2(1 + b)
⇒ –a – 1 = 2 + 2b
⇒ a+ 2b = – 3 … (1)
Also, the point (1, 1) lies on the curve xy + ax + by = 2,we have
1 × 1 + a × 1 + b × 1 = 2
⇒ 1+ a + b = 2
⇒ a+ b = 1 … (2)
From (1) & (2), we get b = -4
Substitute b = – 4 in a + b = 1
a – 4 = 1
⇒ a= 5
So the value of a = 5 & b = – 4
Answer - 3 : -
The given line is x – y + 5 = 0
y = x + 5 is the form of equation of a straight line y = mx + c,where m is the Slope of the line.
So the slope of the line is y = 1 × x + 5
So the Slope is 1. … (2)
Also the point (1, – 6) lie on the tangent, so
x = 1 & y = – 6 satisfies the equation, y = x3 +ax + b
– 6 = 13 + a × 1 + b
⇒ –6 = 1 + a + b
⇒ a+ b = – 7 … (3)
Since, the tangent is parallel to the line, from (1) & (2)
Hence, 3 + a = 1
⇒ a= – 2
From (3)
a + b = – 7
⇒ –2 + b = – 7
⇒ b= – 5
So the value is a = – 2 & b = – 5
Answer - 4 : -
Question - 5 : - Find a point on the curve y = x3 – 2x2 – 2x at which the tangent lines are parallel to the line y = 2x – 3.
Answer - 5 : -
Given the curve y = x3 – 2x2 –2x and a line y = 2x – 3
First, we will find the slope of tangent
y = x3 – 2x2 – 2x
y = 2x – 3 is the form of equation of a straight liney = mx + c, where m is the Slope of the line.
So the slope of the line is y = 2 × (x) – 3
Thus, the Slope = 2. … (2)
From (1) & (2)
⇒ 3x2 –4x – 2 = 2
⇒ 3x2 –4x = 4
⇒ 3x2 –4x – 4 = 0
We will use factorization method to solve the above Quadraticequation.
⇒ 3x2 –6x + 2x – 4 = 0
⇒ 3x (x – 2) + 2 (x – 2) = 0
⇒ (x– 2) (3x + 2) = 0
⇒ (x– 2) = 0 & (3x + 2) = 0
⇒ x= 2 or
x = -2/3
Substitute x = 2 & x = -2/3 in y = x3 –2x2 – 2x
When x = 2
⇒ y= (2)3 – 2 × (2)2 – 2 × (2)
⇒ y= 8 – (2 × 4) – 4
⇒ y= 8 – 8 – 4
⇒ y= – 4
Question - 6 : - Find a point on thecurve y2 = 2x3 at which the Slope of thetangent is 3
Answer - 6 : -
Given the curve y2 = 2x3 and theSlope of tangent is 3
y2 = 2x3
Differentiating the above with respect to x
dy/dx = 0 which is not possible.
So we take x = 2 and substitute it in y2 =2x3, we get
y2 = 2(2)3
y2 = 2 × 8
y2 = 16
y = 4
Thus, the required point is (2, 4)
Question - 7 : - Find a point on the curve x y + 4 = 0 at which the tangents are inclined at an angle of 45o with the x–axis.
Answer - 7 : -
Substitute in xy + 4 = 0, we get
⇒ x(– x) + 4 = 0
⇒ –x2 + 4 = 0
⇒ x2 =4
⇒ x=
2
So when x = 2, y = – 2
And when x = – 2, y = 2
Thus, the points are (2, – 2) & (– 2, 2)
Find a point on thecurve y = x2 where the Slope of the tangent is equal to the x –coordinate of the point.
Answer - 8 : -
Given the curve is y = x2
y = x2
Differentiating the above with respect to x
From (1) & (2), we get,
2x = x
⇒ x= 0.
Substituting this in y = x2, we get,
y = 02
⇒ y= 0
Thus, the required point is (0, 0)
Question - 9 : - At what point onthe circle x2 + y2 – 2x – 4y + 1 = 0, thetangent is parallel to x – axis.
Answer - 9 : -
⇒ –(x – 1) = 0
⇒ x= 1
Substituting x = 1 in x2 + y2 –2x – 4y + 1 = 0, we get,
⇒ 12 +y2 – 2(1) – 4y + 1 = 0
⇒ 1– y2 – 2 – 4y + 1 = 0
⇒ y2 –4y = 0
⇒ y(y – 4) = 0
⇒ y= 0 and y = 4
Thus, the required point is (1, 0) and (1, 4)
At what point ofthe curve y = x2 does the tangent make an angle of 45o withthe x–axis?
Answer - 10 : -
Given the curve is y = x2
Differentiating the above with respect to x
⇒ y= x2
