Application of Derivatives Exercise 16.2 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

1. Tangent and Normal: Geomatical Interpretation
2. Equation of Tangent and Normal
3. Angle of intersection of two curves

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Exercise 16.2

1. Find the slope and the inclination with the x-axis of the tangent of:

a. 2y = 2 – x² at x=1

Solution:

2y = 2 – x²

Differentiating both sides w.r.t. x

Or, 2.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0 – 2x.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = - x.

At x = 1, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -1.

So, slope = -1.

And tanθ = -1.

Where θ is the angle made by the tangent with x – axis.

So, θ = $\frac{{3{\rm{\pi }}}}{4}$.

b. y = - 3x – x⁴ at x = - 1

Solution:

y = - 3x – x⁴

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -3.1 – 4x³ = - 3 – 4x³.

At x = -1, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -3 – 4 * (-1)³ = 1.

So, slope = 1.

And tanθ = 1.

So, θ = $\frac{{\rm{\pi }}}{4}$.

c. x² + y² = 25 at (-3,4)

Solution:

x² + y² = 25

Differentiating both sides w.r.t. x

Or, 2x + 2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -2x.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{\rm{x}}}{{\rm{y}}}$.

At (-3,4), $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{3}{4}$.

So, slope = $\frac{3}{4}$.

And tanθ = $\frac{3}{4}$.

So, θ = tan^-1$\left( {\frac{3}{4}} \right)$.

d. $2x^2+3y^2+8x+3=0$ at (-1,1)

Solution:

2x²+3y²+8x+3=0

Differentiating both sides w.r.t. x

4x + 6y. $\frac{{dy}}{{dx}}$ +8=0

6y. $\frac{{dy}}{{dx}}$ = -8-4x

Slope at (-1,1) is

${\frac{{dy}}{{dx}}_{\left( {-1,1} \right)}} = \frac{{ - 8 - 4x}}{{6y}}$

Or, slope = ${\frac{{dy}}{{dx}}_{\left( { - 1,1} \right)}} = \frac{{ - 8 - 4( - 1)}}{{6(1)}} = \frac{{ - 2}}{3}$

And Angle of inclination: tanθ= $\frac{{ - 2}}{3}$

∴ θ = tan^-1($\frac{{ - 2}}{3}$)

2. At what angle does the curve y(1 + x) = x cut the x-axis ?

Solution:

y(1 + x) = x

The curve meets x – axis where y = 0.

0 = x.

So, x= 0

Or, y = $\frac{{\rm{x}}}{{1 + {\rm{x}}}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\left( {1 + {\rm{x}}} \right).1 - {\rm{x}}.1}}{{{{\left( {1 + {\rm{x}}} \right)}^2}}}$ = $\frac{1}{{{{\left( {1 + {\rm{x}}} \right)}^2}}}$.

At x = 0, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{{{\left( 1 \right)}^2}}}$ = 1.

Tanθ= 1.

So, θ = $\frac{{\rm{\pi }}}{4}$.

 

3. Find the equations of the tangents and normal to the curve

a. y = 2x³ – 5x² + 8 at (2, 4)

Solution:

y = 2x³ – 5x² + 8

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 6x² – 10x → m = $\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)$ = 6.2² = 10.2 = 4.

So, equation of tangent, y – 4 = 4(x – 2) → 4x – y = 4.

And equation of normal, y – 4 = $ - \frac{1}{4}$ (x – 2) → x + 4y = 18.

b. x² – y² = 7 at (4,-3)

Solution:

x² – y² = 7

Differentiating both sides w.r.t. ‘x’.

Or, 2x – 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2x.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{x}}}{{\rm{y}}}$.

At (4,-3)$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{4}{{ - 3}}$ i.e. m = $ - \frac{4}{3}$.

The equation of the tangent at (4,-3) is:

Or, y – (-3) = m(x – 4).

Or, y + 3 = $ - \frac{4}{3}$(x – 4).

Or, 3y + 9 = -4x + 16.

So, 4x + 3y = 7.

The equation of the normal of the normal at (4,-3) is:

Or, y – y₁ = $ - \frac{1}{{\rm{m}}}$(x – x₁).

Or, y – (-3) = $ - \frac{1}{{ - \frac{4}{3}}}$(x – 4).

Or, y + 3 = $\frac{3}{4}$(x – 4).

Or, y + 3 = $\frac{3}{4}$ (x – 4).

Or, 4y + 12 = 3x – 12

So, 3x – 4y = 24.

c) y²=2x at (8,4)

4. Find the points on the curve where the tangents are parallel to the x-axis

a. y = 2x – x²

Solution:

y = 2x – x²

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2.1 – 2x = 2 – 2x.

For tangent parallel to x – axis, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, 2 – 2x = 0

So, x = 1. 

b. y = x³ – 3x² + 1

Solution:

y = x³ – 3x² + 1

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 3x² – 6x + 0 = 3x² – 6x.

For tangent parallel to x – axis, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

So, 3x² – 6x = 0.

Or, 3x(x – 2) = 0

Either, 2x = 0 Or, x = 0

Or, x – 2 = 0

So, x = 0

Or, x – 2 = 0

So, x = 2.

When x = 0, y = x³ – 3x² + 1 = 0 – 0 + 1 = 1.

When x = 2, y = x³ – 3x² + 1 = (2)³ – 3(2)³ + 1 = - 3.

So, the required points are (0,1) and (2,-3). 

c. 4y = x⁴ – 8x²

Solution:

4y = x⁴ – 8x²

Or, 4.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 4x³ – 16x.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = x³ – 4x.

For tangent parallel to x – axis, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, x³ – 4x = 0.

Or, x(x² – 4) = 0

Either, x = 0,

Or, x² = 4.

So, x = ± 2.

When x = 0, 4y = x⁴ – 8x² = 0 – 0.

So, y = 0.

When x = 2,

4y = x⁴ – 8x² = (2)⁴ – 8(2)² = -16.

So, y = -4.

When x = -2,

Or, 4y = x⁴ – 8x² = (-2)⁴ – 8(-2)² = -16.

So, y = -4.

So, the required points are (0,0),(2,-4),(-2,-4). 

5. Find the points on the circle $x ^ 2 + y ^ 2$ = 16 at which the tangents are parallel to the

a) x-axis

b) y-axis

Solution:

x² + y² = 16

Or, 2x + 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0

Or, 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2x.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{\rm{x}}}{{\rm{y}}}$.

For tangent parallel to x –axis, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$  = 0.

Or, $ - \frac{{\rm{x}}}{{\rm{y}}}$ = 0.

So, x = 0.

Or, x² + y² = 16.

Or, 0 + y² = 16.

So, y = ± 4.

So, the points are (0,4) and (0,-4) where the tangents are parallel to x – axis.

Again, for tangent parallel to y – axis, $\frac{{{\rm{dx}}}}{{{\rm{dy}}}}$ = 0.

So, $ - \frac{{\rm{y}}}{{\rm{x}}}$ = 0

So, y = 0.

Or, x² + y² = 16.

Or, x² + 0 = 16.

Or, x² = 16.

So, x = ± 4.

So, the points are (4,0) and (-4,0) where the tangents are parallel to y – axis,

 

6) a. Find the point on the curve $4y = x ^ 2$ where the tangent drawn makes angle 45⁰ with the x-axis.

Solution:

Given curve is 4y = x²…(i)

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}}}{4}$ = $\frac{{\rm{x}}}{2}$ …..(ii)

Let (a,b) be the point on the curve (i) such that, slope at (a,b) is m = $\frac{{\rm{a}}}{2}$  [from(i)]

Since, the tangent at (a,b) makes an angle of 45° with x – axis so, m = 1.

So, $\frac{{\rm{a}}}{2}$ = 1 → a = 2.

Since (a,b) lies on (i) on 4b = a² = 2²

So, b = 1.

So required point = (2,1). 

b. Find the point on the curve $x ^ 2 = 3y + 1$ at which the tangent is parallel to the line 4x + 3y + 5 = 0

Solution:

Given curve is x² = 3y + 1 …(i).

So, 2x = 3.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$→$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}}}{3}$ …(ii)

Let (a,b) be a point on (i) so from (ii), m₁ = $\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)$_(a,b) = $\frac{{2{\rm{a}}}}{3}$.

The given line is 4x + 3y + 5 = 0 ….(iii).

So, slope m₂ = $ - \frac{4}{3}$.

Since the tangent at (a,b) from (i) is parallel to the line (iii) so.

Or, $\frac{{2{\rm{a}}}}{3}$ = $ - \frac{4}{3}$→ a = -2.

Since, (a,b) lies on (i) so a² = 3b + 1.

Or, (-2)² = 3b + 1

So, b = 1

So, required point = (-2,1). 

c. Find the point on the curve $y ^ 2 = 4x + 1$ at which the tangent is perpendicular to the line 7x + 2y = 10.

Solution:

Given curve is y² = 4x + 1 …(i)

So, 2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 4 →$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{2}{{\rm{y}}}$ …(ii)

Let (a,b) be a point on (i) so from (ii),

m₁ = ${\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)_{{\rm{a}},{\rm{b}}}}$= $\frac{2}{{\rm{b}}}$.

The given line is 7x + 2y = 10 …(iii)

So, slope m₂ = $ - \frac{7}{2}$.

Since the tangent at (a,b) for (i) is perpendicular to (iii) so,

Or, m₁.m₂ = -1 →$\frac{2}{{\rm{b}}}. - \frac{7}{2}$ = -1.

So, b = 7.

Since (a,b) lies on (i) so,

Or, b² = 4a + 1

Or, 49 = 4a + 1.

So, a = 12

So, required point is (12,7).

 

7. Show that the equation of the tangent to the curve $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point (a, b) is $\frac{x}{a} + \frac{y}{b} = 2$

Solution:

Given curve is $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1…(i)

So, $\frac{{2{\rm{x}}}}{{{{\rm{a}}^2}}} + \frac{{2{\rm{y}}}}{{{{\rm{b}}^2}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0 →$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{{\rm{b}}{{\rm{x}}^2}}}{{{{\rm{a}}^2}{\rm{y}}}}$.

So, slope = ${\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)_{{\rm{a}}.{\rm{b}}}}$= $\frac{{ - {{\rm{b}}^2}{\rm{a}}}}{{{{\rm{a}}^2}.{\rm{b}}}}$ = $ - \frac{{\rm{b}}}{{\rm{a}}}$.

So, the equation of tangent at (a,b) is:

Or, y – b = $ - \frac{{\rm{b}}}{{\rm{a}}}$ (x – a).

Or, ay – ab = -bx + ab.

Or, bx + ay = 2ab.

So, $\frac{{\rm{x}}}{{\rm{a}}}$ + $\frac{{\rm{y}}}{{\rm{b}}}$ = 2.

 

8. Find the angle of intersection of the curves

a. y = 6 – x² and 2y = x² 

Solution:

Given curve are,

y = 6 – x² …(i) and

2y = x² …(ii)

Thus from (i) and (ii) we have,

Or, y = 6 – 2y → 3y = 6.

So, y = 2.

So from (i), x² = 6 – y = 6 – 2 = 4.

So, x = ± 2.

So, the point of intersection of A(2,2) and B(-2,2).

Now, since y = 6 – x²→$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = -2x.

And y = $\frac{{{{\rm{x}}^2}}}{2}$→$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}}}{2}$ = x.

For A(2,2)

Or, m₁ = - 2 * 2 = -4.

Or, m₂ = 2.

So, θ = tan^-1$\left( {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}.{{\rm{m}}_2}}}} \right)$ = tan^-1$\left( {\frac{{ - 4 - 2}}{{1 - 4.2}}} \right)$ = tan^-1$\left( {\frac{6}{7}} \right)$.

For B(-2,2),

m₁ = -2 * 2 = - 4.

m₂ = -2.

So, θ = tan^-1$\left( {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}.{{\rm{m}}_2}}}} \right)$ = tan^-1$\left( {\frac{{ - 4 - 2}}{{1 - 4.2}}} \right)$ = tan^-1$\left( {\frac{6}{7}} \right)$.

Hence, tan^-1$\left( {\frac{6}{7}} \right)$ at (2,2) and tan^-1$\left( { - \frac{6}{7}} \right)$ at (-2,2). 

b. 4y = x² + 12 and y² = 8x at (2,4)

Solution:

4y = x² + 12

So, 4.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{x}}}{2}$.

At (2,4),m₁ = $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{2}{2}$ = 1.

Again, y² = 8x.

Or, 2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 8.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{4}{{\rm{y}}}$.

At (2,4), m₂ = $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{4}{4}$ = 1.

Tanθ = $\left( {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}.{{\rm{m}}_2}}}} \right)$ = $\frac{{1 - 1}}{{1 + 1.1}}$ = 0.

So, θ = 0.

i.e. the tangents coincide.

So, there is a common tangent at (2,4). 

c. 2y = x² and 2x² y = 1

Solution:

Or, 2y = x² and 2x² y = 1.

Solving the two equations,

Or, x².x² = 1

Or, x⁴ = 1.

So, x = ± 1.

When x = 1, y = $\frac{1}{2}$.

Or, x = -1, y = $\frac{1}{2}$.

The common points are $\left( {1,\frac{1}{2}} \right)$ and $\left( { - 1,\frac{1}{2}} \right)$.

Or, 2y = x².

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}$.2x = x.

At, $\left( {1,\frac{1}{2}} \right)$, m₁ = $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 1.

Again, 2x²y = 1.

Or, 2x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y.4x = 0.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{4{\rm{xy}}}}{{2{{\rm{x}}^2}}}$ = -2$\frac{{\rm{y}}}{{\rm{x}}}$.

At $\left( {1,\frac{1}{2}} \right)$, m₂ = $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - 2$$\frac{{\frac{1}{2}}}{1}$ = -1.

So, m₁ * m₂ = 1 * (-1) = -1.

So, the two tangents at $\left( {1,\frac{1}{2}} \right)$ are orthogonal.

Similarly, the two tangents at $\left( { - 1,\frac{1}{2}} \right)$ are orthogonal.

9. a) Prove that the tangents to the curve $y = x ^ 2 - 3x + 4$ at (1, 2) and (2, 1) are perpendicular to each other.

Solution:

y=x²-3x+4

Differentiating both side with respect to x, we get

$\frac{{dy}}{{dx}} = 2x – 3$

For the point (1,2) the slope of the tangent

Slope m₁ = $[{\frac{{dy}}{{dx}}_{\left( {1,2} \right)}}$=2-3=-1

For the point (2,1) the slope of the tangent

Slope (m₂)= $[{\frac{{dy}}{{dx}}_{\left( {2,1} \right)}}$= 2×2-3=1

Since m₁ ×m₂ = -1, then the tangents to the curve $y = x ^ 2 - 3x + 4$ at (1, 2) and (2, 1) are perpendicular to each other.

b) Prove the tangents to the curve $y = x ^ 3 – 5$ at (1, 5) and (-1, 5) are parallel.

Solution:

y= x³-5

Differentiating both side with respect to x, we get

$\frac{{dy}}{{dx}} = 3{x^2}$

For the point (1,5) the slope of the tangent

Slope m₁ = ${\frac{{dy}}{{dx}}_{at{\rm{ }}(1,5)}}$=3×1²=3

For the point (-1,5) the slope of the tangent

Slope (m₂)= ${\frac{{dy}}{{dx}}_{at{\rm{ }}(-1,5)}}$= 3×(-1)²=3

Since m₁=m₂, then the tangents to the curve y = x³ + 6 at the points (-1,5) and (1,5) are parallel.