# Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.1 | Set 1

### Question 1. Find the slopes of the tangent and the normal to the following curves at the indicated points:

(i) y = âˆšx3 at x = 4

(ii) y = âˆšx at x = 9

(iii) y = x3 â€“ x at x = 2

(iv) y = 2x2 + 3 sin x at x = 0

(v) x = a(Î¸ â€“ sin Î¸), y = a(1 + cos Î¸) at Î¸ = â€“Ï€/2

(vi) x = a cos3 Î¸, y = a sin3 Î¸ at Î¸ = Ï€/4

(vii) x = a(Î¸ â€“ sin Î¸), y = a(1 â€“ cos Î¸) at Î¸ = Ï€/2

(viii) y = (sin 2x + cot x + 2)2 at x = Ï€/2

(ix) x2 + 3y + y2 = 5 at (1, 1)

(x) xy = 6 at (1, 6)

Solution:

We know that the slope of the tangent of a curve is given by dy/dx.

And slope of the normal = â€“1/slope of tangent = â€“1/(dy/dx).

### (i) y = âˆšx3 at x = 4

Differentiating y = âˆšx3 with respect to x, we get,

Slope of tangent = 3x1/2/2

At x = 4, slope of tangent becomes,

Slope of tangent = 3(4)1/2/2 = 3(2)/2 = 3

And slope of normal at x = 4 is â€“1/3.

### (ii) y = âˆšx at x = 9

Differentiating y = âˆšx with respect to x, we get,

Slope of tangent = xâ€“1/2/2

At x = 9, slope of tangent becomes,

Slope of tangent = 9â€“1/2/2 = 1/[(3)(2)] = 1/6

And slope of normal at x = 9 is â€“6.

### (iii) y = x3 â€“ x at x = 2

Differentiating y = x3 â€“ x with respect to x, we get,

Slope of tangent = 3x2 â€“ 1

At x = 2, slope of tangent becomes,

Slope of tangent = 3(2)2 â€“ 1 = 3(4) â€“ 1 = 11

And slope of normal at x = 2 is â€“1/11.

### (iv) y = 2x2 + 3 sin x at x = 0

Differentiating y = 2x2 + 3 sin x with respect to x, we get,

Slope of tangent = 4x + 3 cos x

At x = 0, slope of tangent becomes,

Slope of tangent = 4(0) + 3 cos 0 = 3.

And slope of normal at x = 0 is â€“1/3.

### (v) x = a (Î¸ â€“ sin Î¸), y = a (1 + cos Î¸) at Î¸ = â€“Ï€/2

Differentiating x = a (Î¸ â€“ sin Î¸) with respect to Î¸, we get,

=> dx/dÎ¸ = a (1 â€“ cos Î¸) . . . . (1)

Differentiating y = a (1 + cos Î¸) with respect to Î¸, we get,

=> dy/dÎ¸ = a (â€“sin Î¸) . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent = â€“sin Î¸/(1 â€“ cos Î¸)

At Î¸ = â€“Ï€/2, slope of tangent becomes,

Slope of tangent = â€“sin (â€“Ï€/2)/(1 â€“ cos (â€“Ï€/2))

= 1/(1â€“0)

= 1

And slope of normal at Î¸ = â€“Ï€/2 is â€“1.

### (vi) x = a cos3 Î¸, y = a sin3 Î¸ at Î¸ = Ï€/4

Differentiating x = a cos3 Î¸ with respect to Î¸, we get,

=> dx/dÎ¸ = a [(3cos2 Î¸) (â€“sin Î¸)]

= â€“3a cos2 Î¸ sin Î¸ . . . . (1)

Differentiating y = a sin3 Î¸ with respect to Î¸, we get,

=> dy/dÎ¸ = a [(3sin2 Î¸) (cos Î¸)]

= 3a sin2 Î¸ cos Î¸ . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent == â€“ tan Î¸

At Î¸ = Ï€/4, slope of tangent becomes,

Slope of tangent = â€“ tan (Ï€/4)

= âˆ’1

And slope of normal at Î¸ = Ï€/4 is 1.

### (vii) x = a (Î¸ â€“ sin Î¸), y = a (1 â€“ cos Î¸) at Î¸ = Ï€/2

Differentiating x = a (Î¸ â€“ sin Î¸) with respect to Î¸, we get,

=> dx/dÎ¸ = a (1 â€“ cos Î¸) . . . . (1)

Differentiating y = a (1 â€“ cos Î¸) with respect to Î¸, we get,

=> dy/dÎ¸ = a (sin Î¸) . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent = sin Î¸/(1âˆ’cosÎ¸)

= â€“ tan Î¸

At Î¸ = Ï€/2, slope of tangent becomes,

Slope of tangent = sin Ï€/2/(1âˆ’cos Ï€/2)

= 1/(1âˆ’0)

= 1

And slope of normal at Î¸ = Ï€/2 is âˆ’1.

### (viii) y = (sin 2x + cot x + 2)2 at x = Ï€/2

Differentiating y = (sin 2x + cot x + 2)2 with respect to x, we get,

Slope of tangent = 2 (sin 2x + cot x + 2) (2 cos 2x â€“ cosec2 x)

At x = Ï€/2, slope of tangent becomes,

Slope of tangent = 2 (sin 2(Ï€/2) + cot Ï€/2 + 2) (2 cos 2(Ï€/2) â€“ cosec2 (Ï€/2))

= 2 (0 + 0 + 2) (â€“2 â€“ 1)

= â€“12

And slope of normal at x = Ï€/2 is 1/12.

### (ix) x2 + 3y + y2 = 5 at (1, 1)

Differentiating x2 + 3y + y2 = 5 with respect to x, we get,

=> 2x + 3 (dy/dx) + 2y (dy/dx) = 0

=> 2x + dy/dx (2y+3) = 0

=> Slope of tangent = dy/dx = â€“2x/(2y+3)

At x = 1 and y = 1, slope of tangent becomes,

Slope of tangent = â€“2(1)/[2(1)+3] = â€“2/5

And slope of normal at (1, 1) is 5/2.

### (x) xy = 6 at (1, 6)

Differentiating xy = 6 with respect to x, we get,

=> x (dy/dx) + y = 0

=> Slope of tangent = dy/dx = â€“y/x

At x = 1 and y = 6, slope of tangent becomes,

Slope of tangent = â€“6/1 = â€“6

And slope of normal at (1, 6) is 1/6.

### Question 2. Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.

Solution:

We know that the slope of the tangent of a curve is given by dy/dx.

Differentiating xy + ax + by = 2 with respect to x, we get

=> x (dy/dx) + y + a + b (dy/dx) = 2

=> dy/dx = âˆ’(a+y)/(x+b)

As we are given dy/dx = 2, we get,

=> âˆ’(a+y)/(x+b) = 2

Now at x = 1 and y = 1, we get,

=> âˆ’(a+1)/(1+b) = 2

=> âˆ’a âˆ’ 1 = 2 + 2b

=> a + 2b = â€“3 . . . . (1)

Now the point (1, 1) also lies on the curve, so we have,

=> 1 Ã— 1 + a Ã— 1 + b Ã— 1 = 2

=> 1 + a + b = 2

=> a + b = 1 . . . . (2)

Subtracting (1) from (2), we get,

=> â€“b = 1+3

=> b = â€“4

Putting b = â€“4 in (1), we get,

=> a = 1+4 = 5

Therefore, the value of a is 5 and b is â€“4.

### Question 3. If the tangent to the curve y = x3 + ax + b at (1, â€“6) is parallel to the line x â€“ y + 5 = 0, find a and b.

Solution:

We know that the slope of the tangent of a curve is given by dy/dx.

Differentiating y = x3 + ax + b with respect to x, we get

=> dy/dx = 3x2 + a

Now at x = 1 and y = â€“6, we get,

=> dy/dx = 3(1)2 + a

=> dy/dx = 3 + a . . . . (1)

Now this curve is parallel to the line x â€“ y + 5 = 0.

=> y = x + 5

Therefore slope of the line is 1. So, the slope of the curve will also be 1 as slope of parallel lines are equal. So, from (1), we get,

=> dy/dx = 1 . . . . (2)

From (1) and (2), we get,

=> a + 3 = 1

=> a = â€“2 . . . . (3)

Now at x = 1 and y = â€“6, our curve y = x3 + ax + b becomes,

=> â€“6 = 1 + a + b

=> a + b = â€“7

Using (3), we get,

=> b = â€“7 â€“ (â€“2)

=> b = â€“5

Therefore, the value of a is â€“2 and b is â€“5.

### Question 4. Find a point on the curve y = x3 â€“ 3x where the tangent is parallel to the chord joining (1, â€“ 2) and (2, 2).

Solution:

We are given the coordinates of the chord (1, â€“ 2) and (2, 2).

Therefore, slope of the chord == 4

Given curve is y = x3 â€“ 3x. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 3x2 â€“ 3

As the tangent is parallel to the chord, its slope must be equal to 4.

=> 3x2 â€“ 3 = 4

=> 3x2 = 7

=> x =

Putting value of x in the curve y = x3 â€“ 3x, we get

=> y = x (x2 â€“ 3)

=> y =

=> y =

Therefore, the required point is.

### Question 5. Find a point on the curve y = x3 â€“ 2x2 â€“ 2x at which the tangent lines are parallel to the line y = 2x â€“ 3.

Solution:

Given curve is y = x3 â€“ 2x2 â€“ 2x. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 3x2 â€“ 4x â€“ 2 . . . . (1)

Now this curve is parallel to the line y = 2x â€“ 3 whose slope is 2. So, the slope of the curve will also be 2. So, from (1), we get,

=> 3x2 â€“ 4x â€“ 2 = 2

=> 3x2 â€“ 6x + 2x â€“ 4 = 0

=> 3x (x â€“ 2) + 2 (x â€“ 2) = 0

=> (x â€“ 2) (3x + 2) = 0

=> x = 2 or x = â€“2/3

If x = 2, we get

y = (2)3 â€“ 2 Ã— (2)2 â€“ 2 Ã— (2)

= 8 â€“ 8 â€“ 4

= â€“ 4

And if x = â€“2/3, we get,

y = (â€“2/3)3 â€“ 2 Ã— (â€“2/3)2 â€“ 2 Ã— (â€“2/3)

=

= 4/27

Therefore, (2, â€“4) and (â€“2/3, 4/27) are the required points.

### Question 6. Find a point on the curve y2 = 2x3 at which the slope of the tangent is 3.

Solution:

Given curve is y2 = 2x3. We know that the slope of the tangent of a curve is given by dy/dx.

=> 2y dy/dx = 6x2

=> dy/dx = 3x2/y . . . . (1)

As it is given that the slope of tangent is 3, we get,

=> 3x2/y = 3

=> 3x2 = 3y

=> x2 = y

Putting this in the curve y2 = 2x3, we get,

=> (x2)2 = 2x3

=> x4 âˆ’ 2x3 = 0

=> x3 (x âˆ’ 2) = 0

=> x = 0 or x = 2

If x = 0, we get, y = 0. Putting these values in (1), we get dy/dx = 0, which is not possible as the given value of slope is 3.

And if x =2, we get y = 4.

Therefore, the required point is (2, 4).

### Question 7. Find a point on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45o with the xâ€“axis.

Solution:

Given curve is xy + 4 = 0. We know that the slope of the tangent of a curve is given by dy/dx.

=> x (dy/dx) + y = 0

=> dy/dx = âˆ’y/x . . . . (1)

We are given that the tangent is inclined at an angle of 45o with the xâ€“axis. So slope of the tangent is,

dy/dx = tan 45o = 1.

So, (1) becomes,

=> âˆ’y/x = 1

=> y = âˆ’x

Putting this in the curve xy + 4 = 0 we get,

=> x(âˆ’x) + 4 = 0

=> x2 = 4

=> x = Â±2

When x = 2, y = âˆ’2.

And when x = âˆ’2, y = 2.

Therefore, the required points are (2, â€“ 2) & (â€“ 2, 2).

### Question 8. Find a point on the curve y = x2 where the slope of the tangent is equal to the x â€“ coordinate of the point.

Solution:

Given curve is y = x2. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 2x . . . . (1)

It is given that the slope of the tangent is equal to the x â€“ coordinate of the point.

Therefore, dy/dx = x . . . . (2)

From (1) & (2), we get,

2x = x

=> x = 0

Putting this in the curve y = x2, we get,

=> y = 02

=> y = 0

Therefore, the required point is (0, 0).

### Question 9. At what point on the circle x2 + y2 â€“ 2x â€“ 4y + 1 = 0, the tangent is parallel to x â€“ axis.

Solution:

Given circle is x2 + y2 â€“ 2x â€“ 4y + 1 = 0. We know that the slope of the tangent of a curve is given by dy/dx.

=> 2x + 2y (dy/dx) â€“ 2 â€“ 4 (dy/dx) = 0

=> dy/dx = (1â€“ x)/(yâ€“ 2)

As the tangent is parallel to x-axis, its slope is equal to 0.

So, (1â€“ x)/(yâ€“ 2) = 0

=> x = 1

Putting x = 1 in the circle x2 + y2 â€“ 2x â€“ 4y + 1 = 0, we get,

=> 1 + y2 â€“ 2 â€“ 4y +1 = 0

=> y2 â€“ 4y = 0

=> y (y â€“ 4) = 0

=> y = 0 and y = 4

Therefore, the required points are (1, 0) and (1, 4).

### Question 10. At what point of the curve y = x2 does the tangent make an angle of 45o with the xâ€“axis?

Solution:

Given curve is y = x2. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 2x . . . . (1)

We are given that the tangent is inclined at an angle of 45o with the xâ€“axis. So slope of the tangent is

Therefore, dy/dx = tan 45o = 1 . . . . (2)

From (1) & (2), we get,

2x = 1

=> x = 1/2

Putting this in the curve y = x2, we get,

=> y = (1/2)2

=> y = 1/4

Therefore, the required point is (1/2, 1/4).

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