22.20 - Miscellaneous Examples on Applications of Derivatives - Part 2

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.

Solved example 22.70
Find the equation of the normal to the curve x² = 4y which passes through the point (1,2).
Solution:
1. In the fig.22.71 below, the curve x² = 4y is drawn in red color.

Fig.22.70

2. We have to draw a normal to this curve. That normal should pass through (1,2).
• Substituting x = 1 and y = 2 in the equation x² = 4y, we find that, (1,2) does not lie in the given curve.
• We cannot draw "any line" from (1,2) towards the curve. The line must satisfy the following 2 conditions:
(i) The line must intersect the curve at (h,k)
(ii) The tangent of the curve at (h,k) should be perpendicular to the line.
3. So our first aim is to find h and k.
• The given equation is x² = 4y. This can be written as y = x² / 4
• So dy/dx = x/2. That means, the slope of tangent at any point x on the curve can be obtained using the equation:
slope = x/2
• Then the slope at (h,k) will be h/2
⇒ Slope of the normal passing through (1,2) and (h,k) will be -(2/h)
4. Now, slope of line passing through (1,2) and (h,k) is:
$\rm{\frac{k - 2}{h - 1}}$
5. Equating the results in (3) and (4), we get:
$\rm{\frac{k - 2}{h - 1}~=~\frac{-2}{h}}$
6. In the above step, there are two unknowns h and k. But there is only one equation.
• A second equation can be obtained based on the equation of the curve. Since (h,k) is a point on the curve x² = 4y, we get: h² = 4k
• Substituting this in (5), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{k - 2}{h – 1}}    & {~=~}    &{\frac{-2}{h}}    \\ {~\color{magenta}    2    }    &{\implies}    &{\frac{(h^2 / 4) - 2}{h – 1}}    & {~=~}    &{\frac{-2}{h}}    \\ {~\color{magenta}    3    }    &{\implies}    &{h^3 / 4 ~-~ 2h}    & {~=~}    &{-2h + 2}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{h^3}{4}}    & {~=~}    &{2}    \\ {~\color{magenta}    5    }    &{\implies}    &{h^3}    & {~=~}    &{8}    \\ {~\color{magenta}    6    }    &{\implies}    &{h}    & {~=~}    &{2}    \\ \end{array}$                           
7. Using the equation x² = 4y again, we get:
2² = 4k ⇒ k = 1
8. So (h,k) is (2,1)
• Also, the slope of the normal = −(2/h) = −(2/2) = −1
9. Now, the normal passes through (h,k) and has a slope of −(2/h). The equation of such a line can be obtained as:
y − k = −(2/h)(x − h)
⇒ y − 1 = −1(x − 2)
⇒ y − 1 = −x + 2
⇒ x + y = 3

Solved example 22.71
Find the equation of tangents to the curve
y = cos (x+y), −2π ≤ x ≤ 2π that are parallel to the line x + 2y = 0.
Solution:
1. First we differentiate the given equation

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\cos(x+y)}    \\ {~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-\sin(x+y)[1 + \frac{dy}{dx}]}    \\ {~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-\sin(x+y) ~-~ \frac{dy}{dx} \sin(x+y)}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}~+~\frac{dy}{dx} \sin(x+y)}    & {~=~}    &{-\sin(x+y)}    \\ {~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-\sin(x+y)}{1 ~+~ \sin(x+y)}}    \\ \end{array}$

2. The above result in (1) can be used to find the slope of tangent at any point we want.
• We want those points where slope is same as the slope of the line x + 2y = 0.
    ♦ Slope of this line is -1/2.
• Equating this to the result in (1), we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-\sin(x+y)}{1 ~+~ \sin(x+y)}}    & {~=~}    &{\frac{-1}{2}}    \\ {~\color{magenta}    2    }    &{\implies}    &{\frac{\sin(x+y)}{1 ~+~ \sin(x+y)}}    & {~=~}    &{\frac{1}{2}}    \\ {~\color{magenta}    3    }    &{\implies}    &{2 \sin(x+y)}    & {~=~}    &{1 ~+~ \sin(x+y)}    \\ {~\color{magenta}    4    }    &{\implies}    &{\sin(x+y)}    & {~=~}    &{1}    \\ \end{array}$

3. So we need to solve the equation sin(x+y) = 1
We can apply the first theorem (Details here)
• We know that sin(π/2) = 1.
• So we can write: $\rm{x+y~=~n\pi\,+\,(-1)^n \frac{\pi}{2}}$, where n is any integer.
• This is same as: $\rm{x+y~=~\left(n\,+\,(-1)^n \frac{1}{2}\right)\pi~=~u \pi}$
• Some of the possible values of u are tabulated below:
n = −4 ⇒ u = −3.5
n = −3 ⇒ u = −3.5
n = −2 ⇒ u = −1.5
n = −1 ⇒ u = −1.5
n =    0 ⇒ u = 0.5
n =    1 ⇒ u = 0.5
n =    2 ⇒ u = 2.5
n =    3 ⇒ u = 2.5
n =    4 ⇒ u = 4.5

• The given domain is [−2π,2π]. So the acceptable values of u are: −1.5 and 0.5.
• That means, (x+y) can have two values:
    ♦ −1.5π = −(3/2)π
    ♦    0.5π = (1/2)π.
• We can write two facts:
(i) At the point (x,y), where (x+y) = −(3/2)π, the tangent will be parallel to the given line.
(ii) At the point (x,y), where (x+y) = (1/2)π also, the tangent will be parallel to the given line.

4. We obtained the sum (x+y). If we can find x or y, we will be able to calculate the other.
• Let us calculate y. It can be done in 3 steps:
(i) Consider the two points that we determined in (3). At those two points, cos(x+y) will be zero. This is because, (x+y) is a multiple of (1/2)π.
(ii) So from the given equation of the curve, we get:
y = cos (x+y) = 0
(iii) That means, at both the points determined in (3), y will be zero

5. Now we can calculate the values of x:
• When (x + y) = −(3/2)π, we get (x+0) = −(3/2)π
⇒ x = −(3/2)π 
• When (x + y) = (1/2)π, we get (x+0) = (1/2)π
⇒ x = (1/2)π

6. So the two points are: [−(3/2)π, 0] and [(1/2)π, 0].
We can easily write the equation of the line passing through each of those points. The slope is −(1/2).
• Line through [−(3/2)π, 0]:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-0}    & {~=~}    &{\frac{-1}{2} \left(x - (-3/2)\pi \right)}    \\ {~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{-1}{2} \left(x + \frac{3 \pi}{2} \right)}    \\ {~\color{magenta}    3    }    &{\implies}    &{2y}    & {~=~}    &{-x - \frac{3 \pi}{2}}    \\ {~\color{magenta}    4    }    &{\implies}    &{4y}    & {~=~}    &{-2x - 3 \pi}    \\ {~\color{magenta}    5    }    &{\implies}    &{2x + 4y + 3 \pi}    & {~=~}    &{0}    \\ \end{array}$

• Line through [(1/2)π, 0]:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-0}    & {~=~}    &{\frac{-1}{2} \left(x – (1/2)\pi \right)}    \\ {~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{-1}{2} \left(x + \frac{\pi}{2} \right)}    \\ {~\color{magenta}    3    }    &{\implies}    &{2y}    & {~=~}    &{-x - \frac{\pi}{2}}    \\ {~\color{magenta}    4    }    &{\implies}    &{4y}    & {~=~}    &{-2x - \pi}    \\ {~\color{magenta}    5    }    &{\implies}    &{2x + 4y + \pi}    & {~=~}    &{0}    \\ \end{array}$

7. The graph is shown below:

Fig.22.71

• The given curve is drawn in red color.
• The given line is drawn in green color.
• We see that:
The two tangents (magenta color) are parallel to the green line.

Solved example 22.72
Show that the normal at any point 𝜃 to the curve
x = a cos 𝜃 + a 𝜃 sin 𝜃 , y = a sin 𝜃 − a 𝜃 cos 𝜃
is at a constant distance from the origin.
Solution:
1. First we differentiate the given equation

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{a \cos \theta + a \theta \sin \theta}    \\ {~\color{magenta}    2    }    &{\implies}    &{\frac{dx}{d \theta}}    & {~=~}    &{a (-\sin \theta) + a \theta (\cos \theta) + a (1) \sin \theta}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{a \theta \cos \theta}    \\ {~\color{magenta}    4    }    &{\implies}    &{y}    & {~=~}    &{a \sin \theta - a \theta \cos \theta}    \\ {~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{d \theta}}    & {~=~}    &{a (\cos \theta) - [a \theta (-\sin \theta) + a (1) \cos \theta]}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{a \cos \theta + a \theta \sin \theta - a \cos \theta}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{a \theta \sin \theta}    \\ {~\color{magenta}    8    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{dy}{d \theta} \div \frac{dx}{d \theta}}    \\ {~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \theta \sin \theta}{a \theta \cos \theta}}    \\ {~\color{magenta}    10    }    &{{}}    &{{}}    & {~=~}    &{\tan \theta}    \\ \end{array}$

2. So the slope of tangent at any point 𝜃 is tan 𝜃.
⇒ slope of normal at any point is −(1/tan 𝜃)

3. For any point , the coordinates in terms of x and y can be written as:
[(a cos 𝜃 + a 𝜃 sin 𝜃), (a sin 𝜃 − a 𝜃 cos 𝜃)]

4. Now we have point and slope. The equation of the normal can be obtained as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y – y_1}    & {~=~}    &{m(x – x_1)}    \\ {~\color{magenta}    2    }    &{\implies}    &{y – (a \sin \theta – a \theta \cos \theta)}    & {~=~}    &{\frac{-1}{\tan \theta} \left(x - ( a \cos \theta + a \theta \sin \theta) \right)}    \\ {~\color{magenta}    3    }    &{\implies}    &{y \tan \theta – (a \sin \theta – a \theta \cos \theta) \tan \theta}    & {~=~}    &{\left(- x + ( a \cos \theta + a \theta \sin \theta) \right)}    \\ {~\color{magenta}    4    }    &{\implies}    &{x + y \tan \theta – (a \sin \theta – a \theta \cos \theta) \tan \theta - ( a \cos \theta + a \theta \sin \theta)}    & {~=~}    &{0}    \\ {~\color{magenta}    5    }    &{\implies}    &{x + y \tan \theta – \left[(a \sin \theta – a \theta \cos \theta) \tan \theta + ( a \cos \theta + a \theta \sin \theta) \right]}    & {~=~}    &{0}    \\ \end{array}$

• This is in the form Ax + By + C = 0
Where, A = 1, B = tan 𝜃 and
C = a sin 𝜃 − a 𝜃 cos 𝜃 + a cos 𝜃 + a 𝜃 sin 𝜃 , y =

5. Now we can write the distance:
• Distance d of any point (x₁,y₁) from Ax + By + C = 0, is given by:
$d~=~\frac{\left|A x_1~+~B y_1~+~C \right|}{\sqrt{A^2~+~B^2}}$
• So distance d of origin from Ax + By + C = 0, is given by:
$d~=~\frac{\left|C \right|}{\sqrt{A^2~+~B^2}}$

• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{d}    & {~=~}    &{\frac{(a \sin \theta – a \theta \cos \theta) \tan \theta + ( a \cos \theta + a \theta \sin \theta)}{\sqrt{1 + \tan^2 \theta}}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \frac{\sin^2 \theta}{\cos \theta} – a \theta \sin \theta + a \cos \theta + a \theta \sin \theta}{\sqrt{\sec^2 \theta}}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \frac{\sin^2 \theta}{\cos \theta}+ a \cos \theta }{\sec \theta}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\frac{a \sin^2 \theta ~+~ a \cos^2 \theta}{\cos \theta}}{\sec \theta}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \sin^2 \theta ~+~ a \cos^2 \theta}{\cos \theta} \left(\frac{1}{\sec \theta} \right)}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{a (\sin^2 \theta ~+~ \cos^2 \theta)}{1}}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{a}    \\ \end{array}$

• So the distance is a constant.

6. The graph of the given function is shown in fig.22.72 below. Value of a is assumed as 30.

Fig.22.72

• The graph is drawn in pink color. Two random points are marked in yellow color. The normal at those points are drawn in green color.
• Both the normals have the same distance of 30 from the origin.

Solved example 22.73
The slope of the tangent to the curve
x = t² + 3t −8, y = 2t² − 2t − 5
at the point (2,−1) is
(A) 22/7    (B) 6/7    (C) 7/6    (D) −6/7.
Solution:
1. First we differentiate the given equation
• x = t² + 3t −8
⇒ dx/dt = 2t + 3
• y = 2t² − 2t − 5
⇒ dy/dt = 4t − 2
dy/dx = (dy/dt)(dt/dx) = (4t − 2)/(2t + 3)

2. Given that x = t² + 3t −8
• So for the point (2, −1), we can write:
2 = t² + 3t −8
⇒ t² + 3t −10 = 0
Solving this quadratic equation, we get: t = 2 and t = −5

3. Let us check using y values:
Given that y = 2t² − 2t − 5
• So for the point (2, −1), we can write:
−1 = 2t² − 2t − 5
⇒ 2t² − 2t − 4 = 0
⇒ t² − t − 2 = 0
Solving this quadratic equation, we get: t = 2 and t = −1

4. (t = 2) is common for both x and y. So we can write:
The point (2, −1) is obtained when t = 2

5. So the slope at t = 2 can be written as:
(dy/dx)_{t = 2} = (4(2) − 2)/(2(2) + 3) = (8 − 2)/(4 + 3) = 6/7

So the correct option is (B)

Solved example 22.74
The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is
(A) 1    (B) 2    (C) 3    (D) 1/2.
Solution:
1. Fig.22.73 below, shows the rough sketch of the curve y² = 4x
• We assume that, the line y = mx + 1 touches the curve at (h,k).

Fig.22.73

2. First we differentiate the equation y² = 4x
2y(dy/dx) = 4
⇒ dy/dx = 4/(2y) = 2/y
• So the slope at (h,k) will be: 2/k

3. The line y = mx + 1 is the tangent at (h,k).
So m = 2/k
• Therefore, the equation of the line becomes:
y = (2/k)x + 1

4. Points of intersection of the line and curve can be obtained by solving their equations:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{4x}    \\ {~\color{magenta}    2    }    &{\implies}    &{\left(\frac{2x}{k}~+~1 \right)^2}    & {~=~}    &{4x}    \\ {~\color{magenta}    3    }    &{\implies}    &{\frac{4x^2}{k^2}~+~\frac{4x}{k}~+~1}    & {~=~}    &{4x}    \\ {~\color{magenta}    4    }    &{\implies}    &{4x^2 + 4kx + k^2}    & {~=~}    &{4k^2 x}    \\ {~\color{magenta}    5    }    &{\implies}    &{4x^2 + 4kx – 4k^2 x + k^2}    & {~=~}    &{0}    \\ {~\color{magenta}    6    }    &{\implies}    &{4x^2 + 4k(1 – k)x + k^2}    & {~=~}    &{0}    \\ \end{array}$                           

• For finding the points of intersection, we need to solve the above quadratic equation.
• But since the line is a tangent, there will be only one point of intersection. That means, for the above quadratic equation, there will be only one solution.
• Recall that, if the quadratic equation ax² + bx + c has only one solution, the discriminant b² - 4ac will be zero.
• So for our present quadratic equation, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{[4k(1 – k)]^2 ~-~ 4(4)k^2}    & {~=~}    &{0}    \\ {~\color{magenta}    2    }    &{\implies}    &{16 k^2 (1-k)^2 ~-~ 4(4)k^2}    & {~=~}    &{0}    \\ {~\color{magenta}    3    }    &{\implies}    &{16 k^2 (1-2k + k^2) ~-~ 4(4)k^2}    & {~=~}    &{0}    \\ {~\color{magenta}    4    }    &{\implies}    &{16 k^2 – 32 k^3 + 16 k^4 ~-~ 16 k^2}    & {~=~}    &{0}    \\ {~\color{magenta}    5    }    &{\implies}    &{– 32 k^3 + 16 k^4}    & {~=~}    &{0}    \\ {~\color{magenta}    6    }    &{\implies}    &{– 2 +  k}    & {~=~}    &{0}    \\ {~\color{magenta}    7    }    &{\implies}    &{k}    & {~=~}    &{2}    \\ \end{array}$

5. So from (3), we get:
m = 2/k = 2/2 = 1
• Thus the correct option is (A).

Solved example 22.75
The normal at the point (1,1) on the curve
2y + x² = 3 is
(A) x+y=0    (B) x−y=0    (C) x+y+1=0    (D) x−y=0.
Solution:
1. First we differentiate the equation of the curve. We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2y + x^2}    & {~=~}    &{3}    \\ {~\color{magenta}    2    }    &{\implies}    &{2 \frac{dy}{dx} + 2x}    & {~=~}    &{0}    \\ {~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx} + x}    & {~=~}    &{0}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{-x}    \\ \end{array}$                           
• Using this result, we can find the slope of tangent at any point.

2. The slope of tangent  at (1,1) will be −1.
• So the slope of normal at (1,1) will be the −ve reciprocal, which is 1.

3. The normal has a slope of 1 and it passes through (1,1).
• So the equation of the normal can be obtained as:
y − 1 = 1(x − 1)
⇒ y − 1 = x − 1
⇒ x − y = 0
• So the correct option is (B)

Solved example 22.76
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are
$\rm{(A)~\left(4,\,\pm \frac{8}{3} \right)~~~(B)~\left(4,\,- \frac{8}{3} \right)~~~(C)~\left(4,\,\pm \frac{3}{8} \right)~~~(D)~\left(\pm4,\, \frac{3}{8} \right)}$
Solution:
1. First we differentiate the equation of the curve. We get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{9 y^2}    & {~=~}    &{x^3}    \\ {~\color{magenta}    2    }    &{\implies}    &{9(2y)\frac{dy}{dx}}    & {~=~}    &{3x^2}    \\ {~\color{magenta}    3    }    &{\implies}    &{(6y)\frac{dy}{dx}}    & {~=~}    &{x^2}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x^2}{6y}}    \\ \end{array}$                           

• Using this result, we can find the slope of tangent at any point.
• Therefore, the slope of normal at any point will be given by the −ve reciprocal, which is: $\rm{\frac{-6y}{x^2}}$

2. Let (h,k) be the point on the curve at which, the normal is drawn.
• Then the slope of that normal will be: $\rm{\frac{-6k}{h^2}}$
• So we have a "point on the normal" and the "slope of the normal". Then the equation of the normal can be written as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-k}    & {~=~}    &{\frac{-6k}{h^2}(x-h)}    \\ {~\color{magenta}    2    }    &{\implies}    &{y-k}    & {~=~}    &{\frac{-6kx}{h^2} ~+~\frac{6k}{h}}    \\ {~\color{magenta}    3    }    &{\implies}    &{y}    & {~=~}    &{\frac{-6kx}{h^2} ~+~\frac{6k}{h} ~+~k}    \\ \end{array}$

• This equation is in the form y = mx + c
• So the y-intercept c is $\rm{\frac{6k}{h} ~+~k}$

3. The intercept form of any line is:
$\rm{\frac{x}{a} + \frac{y}{b} = 1}$
Where 'a' and 'b' are the x and y intercepts respectively.
• In our present case, the intercepts are equal. So the equation becomes:
$\rm{\frac{x}{a} + \frac{y}{a} = 1}$
• So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a} + \frac{y}{b}}    & {~=~}    &{1}    \\ {~\color{magenta}    2    }    &{{}}    &{\frac{x}{\frac{6k}{h} ~+~k} + \frac{y}{\frac{6k}{h} ~+~k}}    & {~=~}    &{1}    \\ {~\color{magenta}    3    }    &{\implies}    &{\frac{h}{\frac{6k}{h} ~+~k} + \frac{k}{\frac{6k}{h} ~+~k}}    & {~=~}    &{1}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{h}{\frac{6k + kh}{h}} + \frac{k}{\frac{6k + kh}{h}}}    & {~=~}    &{1}    \\ {~\color{magenta}    5    }    &{\implies}    &{\frac{h^2}{6k + kh}~+~\frac{kh}{6k + kh}}    & {~=~}    &{1}    \\ {~\color{magenta}    6    }    &{\implies}    &{h^2 + kh}    & {~=~}    &{6k + kh}    \\ {~\color{magenta}    7    }    &{\implies}    &{h^2}    & {~=~}    &{6k}    \\ \end{array}$

4. We need one more equation connecting h and k. For that, we can use the equation of the curve. We get: 9k² = h³

5. So the two equations are:
   ♦ h² = 6k
   ♦ 9k² = h³
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{h^2}    & {~=~}    &{6k}    \\ {~\color{magenta}    2    }    &{{}}    &{9 k^2}    & {~=~}    &{h^3}    \\ {~\color{magenta}    3    }    &{\implies}    &{k^2}    & {~=~}    &{\frac{h^3}{9} ~=~\left(\frac{h^2}{6} \right)^2}    \\ {~\color{magenta}    4    }    &{\implies}    &{\frac{h^3}{9}}    & {~=~}    &{\frac{h^4}{36}}    \\ {~\color{magenta}    5    }    &{\implies}    &{h}    & {~=~}    &{\frac{36}{9}}    \\ {~\color{magenta}    6    }    &{\implies}    &{h}    & {~=~}    &{4}    \\ \end{array}$

6. Substituting this value of h in the other equation, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{9k^2}    & {~=~}    &{h^3}    \\ {~\color{magenta}    2    }    &{{}}    &{9 k^2}    & {~=~}    &{4^3}    \\ {~\color{magenta}    3    }    &{\implies}    &{k^2}    & {~=~}    &{\frac{16(4)}{9}}    \\ {~\color{magenta}    4    }    &{\implies}    &{k}    & {~=~}    &{\frac{\pm 4(\pm 2)}{\pm 3}}    \\ {~\color{magenta}    5    }    &{\implies}    &{k}    & {~=~}    &{\pm{\frac{8}{3}}}    \\ \end{array}$                           

7. So the point (h,k) is
option (A): $\rm{\left(4,\,\pm \frac{8}{3} \right)}$

8. Fig.22.74 below shows the graph:

Fig.22.74

• The curve is plotted in red color.
• The green lines are the normals at $\rm{\left(4,\,\pm \frac{8}{3} \right)}$
• We see that, both the green lines have equal x and y intercepts.

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In the next section, we will see a few more examples.

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