22.5 - Solved Examples on Tangents And Normals

In the previous section, we saw Tangents and Normals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 22.18
Find the equation of the tangent to the curve
$\rm{y\,=\,\frac{x-7}{(x-2)(x-3)}}$ at the point where it cuts the x-axis.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

2. We want the point at which the curve cuts the x-axis.
At that point, the y-coordinate will be zero. So in the equation of the curve, we substitute y by zero. We get:
$\rm{0\,=\,\frac{x-7}{(x-2)(x-3)}}$
⇒ x − 7 = 0
⇒ x = 7
• So the required point is: (7,0)

3. Next we want the slope at (7,0). We have:

4. So the equation of the tangent can be written as:
$\rm{y - y_0 ~=~m(x-x_0)}$
⇒ $\rm{y - 0 ~=~\frac{1}{20}(x-7)}$
⇒ 20y = x − 7
⇒ 20y − x + 7 = 0

Fig.22.18

• The graph is shown in fig.22.18 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.

• The tangent is drawn at (7,0)
• The slope triangle has a height of 0.2 units and base of 4 units. So the slope of tangent is 0.2/4 = 1/20

Solved example 22.19
Find the equation of the tangent and normal to the curve
$\rm{x^{2/3} \,+\, y^{2/3}\,=\,2}$ at (1,1).
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

 

2. So we can write the slope of the tangent at (1,1):

$\rm{\left. \frac{dy}{dx} \right|_{(1,1)}~=~(-1) \left(\frac{1}{1} \right)^{1/3}~=~-1}$

3. Now we can write the equation of the tangent at (1,1).
y − y₀ = m(x − x₀)
⇒ y − 1 = (−1)(x − 1)
⇒ y − 1 = −x + 1
⇒ y + x − 2 = 0

4. Slope of the normal is equal to the negative reciprocal of that of the tangent. So slope of the normal is 1.
• Now we can write the equation of the tangent at (1,1).
y − y₀ = m(x − x₀)
⇒ y − 1 = (1)(x − 1)
⇒ y − 1 = x − 1
⇒ y − x = 0

• The graph is shown in fig.22.19 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.

Fig.22.19

• The tangent is drawn at (1,1)
• The slope triangle has a height of −2 units and base of 2 units. So the slope of tangent is −2/2 = −1

Solved example 22.20
Find the equation of the tangent to the curve given by
$\rm{x \,=\, a \sin^3 t,~~y\,=\,b \cos^3 t}$ at  a point where t = π/2.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

2. Next we want the slope at t = π/2. We have:
$\rm{\frac{dy}{dx}\,=\,\frac{-a \sin t}{b \cos t}}$
$\rm{~=\,\frac{-a \sin (\pi/2)}{b \cos (\pi/2)}}$
$\rm{~=\,\frac{-b \cos (\pi/2)}{a \sin (\pi/2)}}$
$\rm{~=\,\frac{-b (0)}{a (1)}}~=~0$

3. Next we want the (x,y) coordinates at t = π/2
$\rm{x \,=\, a \sin^3 (\pi/2)\,=\,a (1)^3 \,=\,a}$
$\rm{y\,=\,b \cos^3 (\pi/2)\,=\,b(0)^3 = 0}$

4. The slope of the tangent is zero. That means, the tangent is horizontal. So we can write the equation of the tangent just by using the y-coordinate obtained in (4).
• We get: y = 0

5. Fig.22.20 below shows the graph of the given function.
• It is assumed that, a = 3 and b = 4
• So the point (a,0) is (3,0)

Fig.22.20

• We see that:
If we draw the tangent at (3,0), it will be same as the x-axis.
• So the equation of the tangent is: y = 0

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The link below gives a few more solved examples:

Exercise 22.3

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In the next section, we will see Approximations.

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