Ncert Solutions For Class 12 Maths Ex 6.3

Ncert Solutions For Class 12 Maths Chapter 6 Ex 6.3

Q-1) x = 3y4 – 4y is a curve find the slope of the tangent to this curve at y = 4.

Ans.)

Here, x = 3y4 – 4y

The slope of the tangent to the curve is

\(\frac{\mathrm{d} x}{\mathrm{d} y} = 12y^{3} – 4\)

Now, at y = 4

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = 12(4)3 – 4 = 12(64) – 4

= 764

Thus, the slope of the tangent to the curve at y = 4 is 764.

Q-2) x = \(\frac{y – 1}{y – 2}; y \neq 2\) is a curve find the slope of the tangent to this curve at y = 4.

Ans.)

Here, x = \(\frac{y – 1}{y – 2}; y \neq 2\)

The slope of the tangent to the curve is

\(\frac{\mathrm{d} x}{\mathrm{d} y} = \frac{(y – 2)(1) – (y – 1)(1)}{(y – 2)^{2}}\)

= \(= \frac{y – 2 – y + 1}{(y – 2)^{2}} = \frac{- 1}{(y – 2)^{2}}\)

Now, y = 4

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = \(\frac{- 1}{(4 – 2)^{2}} = \frac{-1}{4}\)

Thus, the slope of the tangent to the curve at y = 4 is (-1/4).

Q-3) x = y3 – y + 1 is a curve find the slope of the tangent to this curve at a point where y – co-ordinate is 3.

Ans.)

Here, x = 2y3 – y + 1

The slope of the tangent to the curve is

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = 6y2 – 1

Now, at y = 3

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = 6(3)2 – 1 = 6(27) – 1

= 161

Thus, the slope of the tangent to the curve at y = 3 is 161.

Q-4) x = y3 – 4y + 1 is a curve find the slope of the tangent to this curve at a point where y – co-ordinate is 5.

Ans.)

Here, x = y3 – 4y + 1

The slope of the tangent to the curve is

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = 3y2 – 4

Now, at y = 3

\(\frac{\mathrm{d} x}{\mathrm{d} y}\) = 3(5)2 – 4 = 3(25) – 4

= 71

Thus, the slope of the tangent to the curve at y = 5 is 71.

Q-5)x = \(a\cos^{3}t\) and y = \(a\sin^{3}t\) are curve so, find the slope of normal to these curves at t = \(\frac{\pi}{4}\)

Ans.)

Here, x = \(a\cos^{3}t\) and y = \(a\sin^{3}t\)

\( \frac{\mathrm{d} x}{\mathrm{d} t} = 3a\cos ^{2} t (-sin t) = -3a\cos^{2} t \sin t\)

Similarly,

\(\frac{\mathrm{d} y}{\mathrm{d} t} = 3a\sin^{2} t (\cos t)\)

Now,

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}} =\frac{3a\sin^{2} t (\cos t)}{-3a\cos^{2} t \sin t} = – \frac{\sin t}{\ cos t} = -\tan t\)

Thus, the slope of the tangent at t = \(\frac{\pi}{4}\) is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = – \tan \frac{\pi}{4} = – 1\)

Now, the slope of the normal at t = \(\frac{\pi}{4}\) is

\(\frac{- 1}{\frac{\mathrm{d} y}{\mathrm{d} x}} = \frac{-1}{-1} = 1\)

Q-6) x = \(1 – a\sin t\) and y = \(1 – b\cos^{2} t\) are curve so, find the slope of normal to these curves at t = \(\frac{\pi}{2}\)

Ans.)

Here, x = \(1 – a\sin t\) and y = \(1 – b\cos^{2} t\)

\( \frac{\mathrm{d} x}{\mathrm{d} t} = -a\cos t\)

And

\( \frac{\mathrm{d} y}{\mathrm{d} t} = 2b\cos t (- \sin t) = -2b\sin t \cos t\)

Now, \( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}} = \frac{-2b\sin t \cos t}{-a\cos t} = \frac{2b}{a} \sin t\)

Thus, the slope of the tangent at t = \(\frac{\pi}{2}\) is

\( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2b}{a} \sin \frac{\pi}{2} = \frac{2b}{a}\)

Now, the slope of the normal at t = \(\frac{\pi}{4}\) is

\(\frac{-1}{\frac{\mathrm{d} y}{\mathrm{d} x} } = \frac{-1}{\frac{2b}{a}} = – \frac{a}{2b}\)

Q-7) y = x3 – 6x2 -15x +9 is a curve , find points at which the tangent to this curve is parallel to the x- axis.

Ans.)

Here, y = x3 – 6x2 -15x +9

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 3x2 – 12x – 15

= 3(x2 – 4x – 5)

Here, as the tangent is parallel to the x- axis so, its slope is zero.

So,

3(x2 – 4x – 5) = 0

à(x2 – 4x – 5) = 0

à(x – 5)(x + 1) = 0

  • x = 5 or x = – 1

For, x = 3

à(5)3 – 6(5)2 -15(5) +9

= – 91

For, x = -1

à(-1)3 – 6(-1)2 -15(-1) + 9

= 17

Thus, the required points are (5, -91) and (-1, 17).

Q- 8) y = (x – 2)2 is a curve, find points at which the tangent to this curve is parallel to the line joining (2, 0) and (4, 4).

Ans.)

Here, y = (x – 2)2

It is given that the the tangent to the given curve is parallel to the line joining (2, 0) and (4,4). ……………………….(a)

As we know that the parallel lines are having same slope.

So, the slope of the tangent to the curve is

= \(\frac{4 – 0}{4 – 2} = 4/2 = 2\)

Now,

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 2(x-x2)\)

Now, from (a)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 2\)

2(x – 2) = 2

x = 3

For , x = 3

y = (3 – 2)2 = 1

Thus, the required pint is (3, 1).

Q-9) y = x3 – 11x + 5 is a curve. Find the point on this curve where the tangent is y = x – 11.

Ans.)

Here,

y = x3 – 11x + 5

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 3x2– 11…………(a)

Here, y = x – 11 is given tangent to the curve.

By comparing it with y = mx + c, we get where m is slope of tangent

So, the slope of tangent to the given curve is 1.

Now, from (a)

3x2 – 11 = 1

à 3x2 = 12

àx2 = 4

à\(x = \pm 2\)

For, x = -2

y = (-2)3 – 11(-2) + 5 = 19

For, x = 2

y = (2)3 – 11(2) + 5 = – 9

Thus, the required points are (-2, 19) and (2, -9).

Q-10) Find the equation of the all lines with the use of following data:

All the lines are tangent to the curve \(y = \frac{1}{x – 1},\;x\neq 1\)

The slope of the all the lines are same and is -1.

Ans.)

Here, \(y = \frac{1}{x – 1},\;x\neq 1\)

The slope of the tangent to the curve is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-1}{(x – 1)^{2}}\)

The lines are tangent to the given curve and their slope is -1.

So,

\(\frac{-1}{(x – 1)^{2}} = – 1\)

à(x – 1)2 = 1

à(x – 1) = \(\pm 1\)

àx = 2, 0

For, x = 2

y = \(\frac{1}{2-1}\) = 1

For, x = 0

y = \(\frac{1}{0-1}\) = – 1

Thus, the 2 tangents will pass through (0, -1) and (2, 1).

The equation of tangent passing through the point (0, -1) is

y – (-1) = -1(x – 0)

ày + 1 = -x

ày + x + 1 = 0

The equation of tangent passing through the point (2, 1) is

y – 1 = -1(x – 2)

ày + 1 = -x + 2

ày + x – 3 = 0

Thus, the required equations of lines are

y + x + 1 = 0

y + x – 3 = 0

Q-11) Find the equation of the all lines with the use of following data:

All the lines are tangent to the curve \(y = \frac{1}{x – 3},\;x\neq 3\)

The slope of the all the lines are same and is 2.

Ans.)

Here, \(y = \frac{1}{x – 3},\;x\neq 1\)

The slope of the tangent to the curve is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-1}{(x – 3)^{2}}\)

The lines are tangent to the given curve and their slope is 2.

So,

\(\frac{-1}{(x – 3)^{2}} = 2\)

à2(x – 3)2 = -1

à(x – 3)2 = -1/2

Which is not possible so, there is no tangent to the given curve having slope 2.

Q-12)Find the equation of the all lines with the use of following data:

All the lines are tangent to the curve \(y = \frac{1}{x^{2} – 2x + 3}\)

The slope of the all the line is same and is 0.

Ans.)

Here, \(y = \frac{1}{x^{2} – 2x + 3}\)

The slope of the tangent to the curve is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-(2x – 2)}{(x^{2} – 2x + 3)^{2}} = \frac{-2(x – 1)}{(x^{2} – 2x + 3)^{2}}\)

The lines are tangent to the given curve and their slope is 0.

So,

\( \frac{-2(x – 1)}{(x^{2} – 2x + 3)^{2}} = 0\)

à -2(x – 1) = 0

àx = 1

For, x = 1

y = \(\frac{1}{1 – 2 + 3}\) = ½

Thus, the point is (1, ½)

The equation of tangent with slope zero and passing through the point (1, 1/2) is

y – ½ = 0(x – 1)

ày – ½ = 0

ày = ½

Thus, the equation of the tangent (line) is y = ½.

Q-13) \(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\) is a curve. Find the point on this curve where the tangent is

(i) Parallel to y- axis.

(ii) Parallel to x- axis.

Ans.)

Here, \(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\) is a given curve.

Differentiating the given curve w.r.t. ‘x’.

\(\frac{2x}{9} + \frac{2y}{16}*\frac{\mathrm{d} y}{\mathrm{d} x} = 0\)

à\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-16x}{9y}\)

(i) For tangent to the curve, parallel to y- axis, slope of the normal is 0.

So,

Slope of normal = \(\frac{-1}{\frac{-16x}{9y}} = \frac{9y}{16x} = 0\)

ày = 0

Now, for y = 0

à\(\frac{x^{2}}{9} + \frac{(0)^{2}}{16} = 1\)

àx2 = 9

à\(x = \pm 3\)

Thus, the tangent is parallel to y- axis at points (-3, 0) and (3, 0).

(ii) For tangent to the curve, parallel to x- axis, slope of the tangent is 0.

So,

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-16x}{9y} = 0\)

àx = 0

Now,

à\(\frac{(0)^{2}}{9} + \frac{y^{2}}{16} = 1\)

ày2 = 16

à\(y = \pm 4\)

Thus, the tangent is parallel to x- axis at points (-4, 0) and (4, 0).

Q-14) Find the equations of the normal and tangent to the curves given below at the points which are indicated in the questions.

a) y = x2 at (0, 0)

b)y = x3 at (1, 1)

c) y = sin c and x = cot c at c = \(\frac{\pi}{4}\)

d) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)

e) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)

Ans.)

(a) Here, y = x2 at (0, 0)

Differentiating w.r.t. ‘x’

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 2x\)

At (0, 0),

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 2(0) = 0\)

So, the slope of tangent at (0, 0) is 0.

Now, the equation of tangent

y – 0 = 0(x – 0)

ày = 0

Now, the slope of normal to the curve is

= \(\frac{-1}{Slope\;of\;tangent\;at\;(0, 0)} = \frac{-1}{0} = \infty\)

Hence not defined

Thus, the equation for normal is

x = 0.

(b) Here, y = x3

Differentiating w.r.t. ‘x’

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2}\)

At (1, 1),

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 3(1)^2= 3\)

So, the slope of tangent at (1, 1) is 3.

Now, the equation of tangent

y – 1 = 3(x – 1)

ày = 3x – 2

Now, the slope of normal to the curve is

= \(\frac{-1}{Slope\;of\;tangent\;at\;(0, 0)} = \frac{-1}{3}\)

Now, the equation for normal is

y – 1 = (-1/3)(x – 1)

à3y – 3 – 4 = 0

(c) Here, y = sin c and x = cot c

Differentiating w.r.t. ‘c’

\(\frac{\mathrm{d} y}{\mathrm{d} c} = \cos c\) \(\frac{\mathrm{d} x}{\mathrm{d} c} = -\sin c\) \(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} c}}{\frac{\mathrm{d} x}{\mathrm{d} c}} = \frac{\cos c}{-\sin c} = -\cot c\)

At c = \(\frac{\pi}{4}\)

= \(-\cot \frac{\pi}{4} = -1\)

So, the slope of tangent at c = \(\frac{\pi}{4}\)

When c = \(\frac{\pi}{4}\) so, x = \(\frac{\pi}{4}\) and y = \(\frac{\pi}{4}\) is -1.

Now, the equation of the tangent at \((\frac{\pi}{4},\frac{\pi}{4})\)

\(y – \frac{1}{\sqrt{2}} = 1(x – \frac{1}{\sqrt{2}}) \\ \Rightarrow x + y – \frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}}= 0 \\ \Rightarrow x + y – \sqrt{2} = 0\)

Now, the slope of normal to the curve is

= \(\frac{-1}{Slope\;of\;tangent\;at\;(\frac{\pi}{4},\frac{\pi}{4})} = \frac{-1}{-1} = 1\)

Now, the equation for normal is

= \(y – \frac{1}{\sqrt{2}} = 1(x – \frac{1}{\sqrt{2}}) \\ \Rightarrow x = y\)

(d) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)

Differentiating w.r.t. ‘x’

\(\frac{\mathrm{d} y}{\mathrm{d} x}\) = 4x3 – 18x2 + 26x – 10

At (0, 5),

\(\frac{\mathrm{d} y}{\mathrm{d} x}\) = 4(0)3 – 18(0)2 + 26(0) – 10 = -10

So, the slope of tangent at (0, 5) is -10.

Now, the equation of tangent

y – 5 = -10(x – 0)

à10x + y – 5 = 0

Now, the slope of normal to the curve is

= \(\frac{-1}{Slope\;of\;tangent\;at\;(0, 5)} = \frac{-1}{-10} = \frac{1}{10}\)

Now, the equation for normal is

y – 5 = (1/10)(x – 0)

à10y – 50 = x

àx – 10y = – 50

(e) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)

Differentiating w.r.t. ‘x’

\(\frac{\mathrm{d} y}{\mathrm{d} x}\) = 4x3 – 18x2 + 26x – 10

At (1, 3),

\(\frac{\mathrm{d} y}{\mathrm{d} x}\) = 4(1)3 – 18(1)2 + 26(1) – 10 = 2

So, the slope of tangent at (1, 3) is 2.

Now, the equation of tangent

y – 3= 2(x – 1)

à2x – y +1 = 0

Now, the slope of normal to the curve is

= \(\frac{-1}{Slope\;of\;tangent\;at\;(1, 3)} = \frac{-1}{2}\)

Now, the equation for normal is

y – 3= (-1/2)(x – 1)

à2y – 6 = – x + 1

àx + 2y = 7

Q-15) y = x2 – 2x + 7 is a curve. Find the equation to the tangent line to the given curve which is:

(i) Perpendicular to line -15x + 15y -10 = 0.

(ii) Parallel to line 2x – y + 9 = 0.

Ans:

Here, y = x2 – 2x + 7.

Differentiating w.r.t. ‘x’,

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 2x – 2……………..(a)

(i) The given curve is perpendicular to the line -15x + 5y = 13.

i.e. y = 3x + 13/5

By comparing it with y = mx + c, we get

Slope of the line = 3

The given curve is perpendicular to line y = 3x + 13/5, so the slope of the curve is given by,

\(\Rightarrow \frac{-1}{slope\;of\;the\;line} = \frac{-1}{3}\)

à2x – 2 = -(1/3)

à2x = 2 – (1/3)

à x = 5/6

Now, for x = 5/6,

y = (5/6)2 – 2(5/6) + 7 = 217/36

Thus, the tangent is passing through the point (5/6, 217/36), so its equation is given by

\(y – \frac{217}{36} = \frac{-1}{3}(x – \frac{5}{6})\)

à\(\frac{36y – 217}{36} = \frac{-1}{18}(6x – 5)\)

à36y – 217 = -2((6x – 5)

à36y – 217 = -12x + 10

à36y – 12x – 227 = 0

Hence the required equation of tangent is 36y – 12x – 227 = 0.

(ii) The given curve is parallel to the line 2x – y + 9 = 0.

i.e. y = 2x + 9

By, comparing it with y = mx + c, we get

Slope of the line = 2

As the given curve is parallel to line y = 2x + 9. So, their slope will be equal.

Now, from (a)

2x – 2 = 2

àx = 2

For, x = 2,

y = (2)2 – 2(2) + 7= 7

Thus, the tangent is passing through the point (2, 7), so its equation is given by

y – 7 = 2(x – 2)

ày = 2x – 3 = 0

Hence the required equation of tangent is y = 2x – 3 = 0.

Q-16) y = 7x3 + 11 is a curve, show that the tangent to this curve at x = -2 and x = 2 are parallel.

Ans.)

Here, y = 7x3 + 11

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 21x2

Thus, the slope of tangent to the curve is 21x2.

Now, the slope of the tangent at x = -2 is

= 21(-2)2 = 21*4 = 84

The slope of the tangent at x = 2 is

= 21(2)2 = 21*4 = 84

Here the slope of both the tangent is equal. So, the two tangents are parallel.

Q-17) y = x3 is a curve, find the points on the curve at which the slope of the tangent is equal to the y –co-ordinate of the point.

Ans.)

Here, y = x3

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 3x2

Thus, the slope of tangent to the curve is 3x2.

Now, the slope of the tangent at (x, y) is

= 3(x)2

Now it is given that slope of the tangent is equal to the y –co-ordinate of the point.

\( 3x^{2} = y\) \( 3x^{2} = x^{3}\) \( x^{2}(x – 3) = 0\)

\( \) x = 3, x = 0.

For, x = 0

y = (0)3 = 0

For, x = 3

y = (3)3 = 9

Thus, the required points are (0, 0) and (3, 9).

Q-18) y = 4x3 – 2x5is a curve, find all the points at which the tangent passes through (0, 0).

Ans.)

Here, y = 4x3 – 2x5

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 12x2 – 10x4

Thus, the slope of tangent to the curve is 12x2 – 10x4.

Now, the slope of the tangent at (x, y) is

Y – y = (12x2 – 10x4)(X – x)……………………………(a)

Here, the tangent passes through (0,0), So X = 0 and Y = 0.

Thus, from (a)

-y = (12x2 – 10x4)(-x)

ày = 12x3 – 10x5

à4x3 – 2x5 = 12x3 – 10x5

à8x5 – 8x3 = 0

àx5 – x3 = 0

àx3(x2 – 1)= 0

à\(x = 0 \;or\;x = \pm 1\)

Now, for x = 0

y = 4(0)3 – 2(0)5 = 0

For, x = -1

y = 4(-1)3 – 2(-1)5= -2

For, x = 1

y = 4(1)3 – 2(1)5 = 2

Thus, the required points are (0,0), (-1,-2) and (1,2).

Q-19) x2 +y2 -2x -3 = 0 is a curve. Find the points at which the tangent is parallel to x- axis.

Ans.)

Here, x2 +y2 -2x -3 = 0

Differentiating w.r.t. ‘x’,

\(2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} – 2 = 0 \\ \Rightarrow y\frac{\mathrm{d} y}{\mathrm{d} x} = 1 – x \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1 – x}{y}\)

Here, as the tangent is parallel to the x- axis. So, the slope of the tangent is 0.

\( \frac{1 – x}{y} = 0 \\ \Rightarrow 1 – x = 0 \\ \Rightarrow x = 1\)

For, x = 1

(1)2 + y2 -2(1) -3 = 0

ày2 = 4

à\(y = \pm 2\)

Thus, the required points are (1, -2) and (1,2).

Q-20) ty2 = x3 is a curve. Find the equation of normal to this curve at (tn2, tn3).

Ans.)

Here, ty2 = x3

Differentiating w.r.t. ‘x’

\(2ty\frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3x^{2}}{2ty}\)

The slope of tangent at (tn2, tn3) is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3(tn^{2})^{2}}{2t(tn^{3})} = \frac{3t^{2}n^{4}}{2t^{2}n^{3}} = \frac{3n}{2}\)

Now, the slope of normal at (tn2, tn3) is given by

= \(\frac{-1}{Slope\;of\;tangent\;at\;(tn^{2},tn^{3})} = \frac{-2}{3n}\)

Thus, the equation of the normal at(tn2, tn3) is

y – tn3 = \(\frac{-2}{3n}(x – tn^{2})\)

à3ny – 3tn4 = -2x + 2n2

à2x + 3ny – tn2(2 + 3n2) = 0

Q-21) y = x3 + 2x + 6 is a curve. Find the equation of normal to this curve and the normal is parallel to the lone x + 14y + 4 = 0.

Ans.)

Here, y = x3 + 2x + 6

\( \frac{\mathrm{d} y}{\mathrm{d} x}\) = 3x3 + 2

Now, the slope of tangent at(x, y) is

=3x3 + 2

Thus, the slope of normal at (x, y) is given by

= \(\frac{-1}{Slope\;of\;tangent\;at\;(x,y)} = \frac{-1}{3x^{2} + 2}\)

The normal to the curve is parallel to the line x + 14y + 4 = 0, so slope for both of them is equal.

Thus, the slope of line x + 14y + 4 = 0 is \(\frac{-1}{3x^{2} + 2}\)………..(a)

Now, x + 14y + 4 = 0 can be written as

y = – (x/14) – (7/2)

On comparing above equation with y = mx + c, we get

Slope of line x + 14y + 4 = 0 equals to (-1/14).

From (a)

\( \frac{-1}{3x^{2} + 2} = \frac{-1}{14}\)

à3x2 + 2 = 14

à3x2 = 12

àx2 = 4

àx = -2 or x = 2

For, x = -2

y = (-2)3 + 2(-2) + 6 = -6

For, x = 2

y = (2)3 + 2(2) + 6 = 18

Thus, there exits 2 normal to the given curve with same slope of (-1/14) and pasing through two distinct points (-2, -16) and (2,18).

The equation of normal passing through (-2, -16) is

y – (-6) = (-1/14)(x – (-2))

ày +6 = (-1/14)(x + 2)

à14y + 84= -x – 2

àx +14y = -86

The equation of normal passing through (2, 18) is

y – 18 = (-1/14)(x – 2)

à14y – 252 = -x + 2

à x + 14y = 254

Thus, the required equations of the normal are

àx +14y = -86

àx +14y = 254

Q-22) Parabola y2 = 4mxis given. Find the equations for normal and tangent to this parabola at (mt2, 2mt).

Ans.)

Here, y2 = 4mx

Differentiating w.r.t. ‘x’

\(2y\frac{\mathrm{d} y}{\mathrm{d} x} = 4m\)

à\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2m}{y}\)

Now, the slope of tangent at(mt2, 2mt) is

= 2m/2mt = 1/t

Now, the equation of tangent to the given parabola at (mt2, 2mt) is given by,

y – 2mt = 1/t(x – mt2)

àty – 2mt2 = x – mt2

àty = x + mt2

Now, the slope of normal to the parabola at (mt2, 2mt) is given by,

\(\frac{-1}{Slope\;of\;tangent\;at\;(mt^{2},2mt)} = -t\)

Thus, the equation of the normal to parabola at (mt2, 2mt) is

y – 2mt = -t(x – mt2)

ày – 2mt = -tx + mt3

ày = -tx + 2mt + mt3

Q-23)Show that the curves xy = m and x = y2 cut at 90 degrees if 8m2 = 1.

[ Hint: 2 curves intersect at 90 degrees if the tangent to those 2 curves are perpendicular to each other at the point of intersection.]

Ans.)

Here, xy = m and x = y2

à(y2)y = m

ày3 = m

ày = m1/3

Thud, x = m2/3

Hence the point of intersection of curves is (m2/3, m1/3)

Now, x = y2

Differentiating w.r.t. ‘x’

\(1 = 2y\frac{\mathrm{d} y}{\mathrm{d} x} \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2y}\)

Now, slope of tangent to the curve x = y2at (m2/3, m1/3) is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2m^{1/3}}\)

Now, xy = m

Differentiating w.r.t. ‘x’

\(x\frac{\mathrm{d} y}{\mathrm{d} x} + y = 0 \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-y}{x}\)

Now, slope of tangent to the curve xy = m at (m2/3, m1/3) is

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-m^{1/3}}{m^{2/3}} = \frac{-1}{m^{1/3}}\)

As we know that, 2 curves intersect at 90 degrees if the tangent to those 2 curves are perpendicular to each otherat the point of intersection.

i.e. the product of slope of both the tangent is -1.

So, \((\frac{1}{2m^{2/3}})(\frac{-1}{m^{1/3}}) = -1\)

à2m2/3 = 1

à(2m2/3)3 = (1)3

à8m2 = 1

Hence proved.

Q-24) Hyperbola \(\frac{x^{2}}{a^{2}} – \frac{y^{2}}{b^{2}} = 1\) is given. Find the equations of tangent and normal to this hyperbola at point (m, n).

Ans.)

Here, \(\frac{x^{2}}{a^{2}} – \frac{y^{2}}{b^{2}} = 1\)

Differentiating w.r.t. ‘x’

\(\frac{2x}{a^{2}} – \frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x} = 0 \\ \Rightarrow \frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2x}{a^{2}} \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{b^{2}x}{a^{2}y}\)

Now, the slope of tangent at (m, n) is

= \(\frac{b^{2}m}{a^{2}n}\)

The equation of tangent at (m, n) is

\(y – n = \frac{b^{2}m}{a^{2}n}(x – m)\) \(\Rightarrow a^{2}yn – a^{2}n^{2} = b^{2}xm – b^{2}m^{2} \\ \Rightarrow b^{2}xm – b^{2}m^{2} – a^{2}yn + a^{2}n^{2} = 0 \\\)

\(\Rightarrow \frac{xm}{a^{2}} – \frac{yn}{b^{2}} – (\frac{m^{2}}{a^{2}} – \frac{n^{2}}{b^{2}}) = 0\) (dividing both sides by a2b2)

\(\Rightarrow \frac{xm}{a^{2}} – \frac{yn}{b^{2}} – 1 = 0\)

( as (m, n) lies on the hyperbola)

\(\Rightarrow \frac{xm}{a^{2}} – \frac{yn}{b^{2}} = 1\)

The slope of normal at (m, n) is given by,

\(\frac{-1}{Slope\;of\;tangent\;at\;(m,n)} = \frac{-a^{2}n}{b^{2}m}\)

Now, the equation of normal at (m, n) is

\(y – n = \frac{-a^{2}n}{b^{2}m}(x – m) \\ \Rightarrow \frac{y – n}{a^{2}n} = \frac{-(x – m)}{b^{2}m} = 0 \\ \Rightarrow \frac{y – n}{a^{2}n} + \frac{x – m}{b^{2}m} = 0\)

Q-25)\(y = \sqrt{3x – 2}\) is a curve. Find the equations of the tangent to the curve which is parallel to 4x – 2y+ 5 = 0.

Ans.)

Here, \(y = \sqrt{3x – 2}\)

\( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{3}{2\sqrt{3x – 2}}\)

Now, the slope of the tangent at (x, y) is

= \(\frac{3}{2\sqrt{3x – 2}}\)

Now, 4x – 2y + 5 = 0 can be written as

y = 2x + 5/2

Comparing y = 2x + 5/2 with y = mx + c , we get

Slope of line = 2

Now the tangent is parallel to the line 4x – 2y + 5 = 0. So,

Slope of the tangent = Slope of line 4x – 2y+ 5 = 0

\( \frac{3}{2\sqrt{3x – 2}} = 2 \\ \Rightarrow \sqrt{3x – 2} = 3/4 \\ \Rightarrow 3x – 2 = 9/16 \\ \Rightarrow 3x = 9/16 + 2 = 41/6 \\ \Rightarrow x = 41/48\)

For, x = 41/48

\(y = \sqrt{3(\frac{41}{48}) – 2} = \sqrt{\frac{41}{16} – 2} = \sqrt{\frac{41 – 32}{16}} = \sqrt{\frac{9}{16}} = 3/4\)

Now, the equation of tangent at \((\frac{41}{48}, \frac{3}{4})\) is

y – ¾= 2(x -41/48)

\(y – \frac{3}{4} = 2(x – \frac{41}{48}) \\ \Rightarrow \frac{4y – 3}{4} = 2 \\ \Rightarrow 4y – 3 = \frac{48x – 41}{6} \\ \Rightarrow 24y – 18 = 48x – 41 \\ \Rightarrow 48x – 24y = 23\)

Thus, the required equation is 48x -24y -23 = 0.

Q-26) \(y = 2x^{3} + 4\sin x\) is a curve. Find the slope of normal to the curve at x = 0.

(i) -4

(ii) -1/4

(iii) 1/4

(iv) 4

Ans.)

(ii) -1/4

Explanation:

Here, \(y = 2x^{3} + 4\sin x\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = 6x + 4\cos x\)

Now the slope of tangent at x = 0 is

= \(6(0) + 4\cos (0) = 4\)

Now, the slope of the normal to the curve is given by

= \(\frac{-1}{Slope\;of\;tangent\;at\;x = 1} = \frac{-1}{4}\)

Q-27) The line x –y + 1 = 0 is a tangent to the parabola y2 = 4x at point ____.

(i) (2, 1)

(ii) (1, -2)

(iii) (1, 2)

(iv) (-1, 2)

Ans.)

(iii) (1, 2)

Explanation:

Here, y2 = 4x

For finding the slope of tangent

Differentiating w.r.t. ‘x’

\(2y\frac{\mathrm{d} y}{\mathrm{d} x} = 4 \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2}{y}\)……………..(a)

Here, the tangent to the curve is x – y + 1 = 0, which can be written as y = x + 1 and comparing it with y = mx + c, we get

Slope of tangent = 1

From (a)

2/y = 1

ày = 2

For y = 2

2 = x + 1

àx = 1

Thus, the required point is (1, 2).

 

Exercise 4 :

Q-1) Find the approximate value up to 3 decimal places using differentials

(a)\((32.15)^{\frac{1}{5}}\)

(b) \((3.968)^{\frac{3}{2}}\)

(c) \((81.5)^{\frac{1}{4}}\)

(d) \((26.57)^{\frac{1}{3}}\)

(e) \(\sqrt{0.0037}\)

(f) \((401)^{\frac{1}{2}}\)

(g) \((82)^{\frac{1}{4}}\)

(h) \((255)^{\frac{1}{4}}\)

Ans:

(a) \((32.15)^{\frac{1}{5}}\)

Let, \(y = x^{\frac{1}{5}}\).

Let x = 32 and \(\Delta x = 0.15\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{5}} – x^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} – 32^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} – 2 \\ \Rightarrow (32.15)^{\frac{1}{5}} = 2 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{5(x)^{\frac{4}{5}}}.(\Delta x)\) (as \(y = x^{\frac{1}{5}}\))

\(= \frac{1}{5*(2)^{4}}.(0.15) \\ = \frac{0.15}{80} = 0.00187\)

Thus, the approximate value of \((32.15)^{\frac{1}{5}}\) is 2 + 0.00187 = 2.00187.

(b) \((3.968)^{\frac{3}{2}}\)

Let, \(y = x^{\frac{3}{2}}\).

Let x = 4 and \(\Delta x = -0.032\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{3}{2}} – x^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} – 4^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} – 8 \\ \Rightarrow (3.968)^{\frac{3}{2}} = 8 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{3}{2}(x)^{\frac{1}{2}}.(\Delta x)\) (as \(y = x^{\frac{3}{2}}\))

= 1.5(2)(-0.032)

= -0.096

Thus, the approximate value of \((3.968)^{\frac{3}{2}}\) is 8 -0.096 = 7.904.

(c) \((81.5)^{\frac{1}{4}}\)

Let, \(y = x^{\frac{1}{4}}\).

Let x = 81 and \(\Delta x = 0.5\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{4}} – x^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} – 81^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} – 3 \\ \Rightarrow (81.5)^{\frac{1}{4}} = 3 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{4*(x)^{\frac{3}{4}}}.(\Delta x)\) (as \(y = x^{\frac{1}{4}}\))

\(= \frac{1}{4*27}.(0.5) \\ = \frac{0.5}{108} = 0.0046\)

Thus, the approximate value of \((81.5)^{\frac{1}{4}}\) is 3 + 0.0046 = 3.0046.

(d) \((26.57)^{\frac{1}{3}}\)

Let, \(y = x^{\frac{1}{3}}\).

Let x = 27 and \(\Delta x = -0.43\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{3}} – x^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} – 27^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} – 3 \\ \Rightarrow (26.57)^{\frac{1}{3}} = 3 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{3*(x)^{\frac{2}{3}}}.(\Delta x)\) (as \(y = x^{\frac{1}{3}}\))

\(= \frac{1}{3*(9)}(-0.43) \\ = \frac{-0.43}{27} = -0.015\)

Thus, the approximate value of \((26.57)^{\frac{1}{3}}\) is 3 + -0.015 = 2.985.

(e) \(\sqrt{0.0037}\)

Let, \(y = x^{\frac{1}{2}}\).

Let x = 0.0036 and \(\Delta x = 0.0001\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{2}} – x^{\frac{1}{2}} = (0.0037)^{\frac{1}{2}} – (0.0036)^{\frac{1}{2}} = (0.0037)^{\frac{1}{2}} – 0.06 \\ \Rightarrow (0.0037)^{\frac{1}{2}} = 0.06 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{2*(x)^{\frac{1}{2}}}.(\Delta x)\) (as \(y = x^{\frac{1}{2}}\))

\(= \frac{1}{2*(0.06)}(0.0001) \\ = \frac{0.0001}{0.12} = 0.00083\)

Thus, the approximate value of \((0.0037)^{\frac{1}{2}}\) is 0.06 + 0.00083 = 0.06083.

(f) \((401)^{\frac{1}{2}}\)

Let, \(y = x^{\frac{1}{2}}\).

Let x = 400 and \(\Delta x = 1\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{2}} – x^{\frac{1}{2}} = (401)^{\frac{1}{2}} – (400)^{\frac{1}{2}} = (401)^{\frac{1}{2}} – 20 \\ \Rightarrow (401)^{\frac{1}{2}} = 20 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{2*(x)^{\frac{1}{2}}}.(\Delta x)\) (as \(y = x^{\frac{1}{2}}\))

\(= \frac{1}{2*(20)}(1) \\ = \frac{1}{40} = 0.025\)

Thus, the approximate value of \((401)^{\frac{1}{2}}\) is 20 + 0.025 = 20.025.

(g) \((82)^{\frac{1}{4}}\)

Let, \(y = x^{\frac{1}{4}}\).

Let x = 81 and \(\Delta x = 1\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{4}} – x^{\frac{1}{4}} = (82)^{\frac{1}{4}} – 81^{\frac{1}{4}} = (82)^{\frac{1}{4}} – 3 \\ \Rightarrow (82)^{\frac{1}{4}} = 3 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{4*(x)^{\frac{3}{4}}}.(\Delta x)\) (as \(y = x^{\frac{1}{4}}\))

\(= \frac{1}{4*27}.(1) \\ = \frac{1}{108} = 0.009\)

Thus, the approximate value of \((82)^{\frac{1}{4}}\) is 3 + 0.009 = 3.009.

(h) \((255)^{\frac{1}{4}}\)

Let, \(y = x^{\frac{1}{4}}\).

Let x = 256 and \(\Delta x = -1\)

Now,

\(\Delta y = (x + \Delta x)^{\frac{1}{4}} – x^{\frac{1}{4}} = (255)^{\frac{1}{4}} – 256^{\frac{1}{4}} = (255)^{\frac{1}{4}} – 4 \\ \Rightarrow (255)^{\frac{1}{4}} = 5 + \Delta y\)

Also, dy is approximately equal to \(\Delta y\), so we get

\(dy = \frac{\mathrm{d} y}{\mathrm{d} x}\Delta x = \frac{1}{4*(x)^{\frac{3}{4}}}.(\Delta x)\) (as \(y = x^{\frac{1}{4}}\))

\(= \frac{1}{4*256^{\frac{3}{4}}}.(-1) \\ = \frac{-1}{4*64} = -0.0039\)

Thus, the approximate value of \((255)^{\frac{1}{4}}\) is 4 – 0.0039 = 3.9961.

 

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