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.
In the next section, we will see a few more examples.
Previous
Contents
Next
Copyright©2024 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment