Question

Find the equation of the tangent and normal to the following curve
\[x = \dfrac{1}{t}\], \[y = t - \dfrac{1}{t}\] at \[t = 2\].

Answer 1

Hint:
Here, we will find the slope of the tangent using the given information. Then, we will find the point of contact of the tangent to the curve. Substituting these values in the formula of the equation of a tangent, we will be able to find the required equation of the tangent. We will find the slope of normal using the same formula to find the required equation of the normal.

Formula Used:
We will use the following formulas:
1) \[\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}\]
2) Equation of a straight line is \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\]
\[{m_t} \times {m_n} = - 1\] where, \[{m_t}\] and \[{m_n}\] are the slopes of the tangent and the normal respectively.

Complete step by step solution:
We are given a curve where,
\[x = \dfrac{1}{t}\], \[y = t - \dfrac{1}{t}\]
Now, the point of contact of the tangent and this curve be \[\left( {{x_1},{y_1}} \right)\] at Now, since the point of contact lies on the curve as well, thus, it will satisfy its equation and points. Thus,
\[{x_1} = \dfrac{1}{t}\] and \[{y_1} = t - \dfrac{1}{t}\]
Here, substituting \[t = 2\], we get,
\[ \Rightarrow {x_1} = \dfrac{1}{2}\] and \[{y_1} = 2 - \dfrac{1}{2} = \dfrac{{4 - 1}}{2} = \dfrac{3}{2}\]
Therefore, the coordinates of point of contact, \[\left( {{x_1},{y_1}} \right) = \left( {\dfrac{1}{2},\dfrac{3}{2}} \right)\]
Now, we will differentiate the given curve with respect to \[t\]
Hence, we get,
\[\dfrac{{dx}}{{dt}} = \dfrac{{ - 1}}{{{t^2}}}\] and \[\dfrac{{dy}}{{dt}} = 1 + \dfrac{1}{{{t^2}}}\]
Here, we have used the formula \[\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}\]
Now, substituting \[t = 2\]
\[ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{ - 1}}{4}\] and \[\dfrac{{dy}}{{dt}} = 1 + \dfrac{1}{4} = \dfrac{5}{4}\]
Now, dividing \[\dfrac{{dy}}{{dt}}\] by \[\dfrac{{dx}}{{dt}}\], we get,
\[{\dfrac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = \dfrac{{\dfrac{5}{4}}}{{\dfrac{{ - 1}}{4}}} = - 5\]
Therefore, the slope of the tangent \[{m_t}\] is \[ - 5\]
Hence, equation of tangent will be:
\[\left( {y - {y_1}} \right) = {m_t}\left( {x - {x_1}} \right)\]
Here, substituting \[{m_t} = - 5\] and \[\left( {{x_1},{y_1}} \right) = \left( {\dfrac{1}{2},\dfrac{3}{2}} \right)\], we get,
\[ \Rightarrow \left( {y - \dfrac{3}{2}} \right) = - 5\left( {x - \dfrac{1}{2}} \right)\]
Taking LCM inside the brackets on both the sides, we get
\[ \Rightarrow \left( {\dfrac{{2y - 3}}{2}} \right) = - 5\left( {\dfrac{{2x - 1}}{2}} \right)\]
Cancelling out the denominator and solving further, we get
\[ \Rightarrow 2y - 3 = - 10x + 5\]
\[ \Rightarrow 10x + 2y - 8 = 0\]
Dividing both sides by 2, we get
\[ \Rightarrow 5x + y - 4 = 0\]
Therefore, the equation of tangent to the given curve is \[5x + y - 4 = 0\]
Now, tangent is perpendicular to the normal.
Therefore, the product of their slopes will be equal to \[ - 1\].
\[{m_t} \times {m_n} = - 1\]
\[ \Rightarrow - 5 \times {m_n} = - 1\]
Dividing both sides by -5, we get,
\[ \Rightarrow {m_n} = \dfrac{{ - 1}}{{ - 5}} = \dfrac{1}{5}\]
Therefore, the slope of normal \[{m_n} = \dfrac{1}{5}\]
Hence, equation of normal will be:
\[\left( {y - {y_1}} \right) = {m_t}\left( {x - {x_1}} \right)\]
Here, substituting \[{m_n} = \dfrac{1}{5}\] and \[\left( {{x_1},{y_1}} \right) = \left( {\dfrac{1}{2},\dfrac{3}{2}} \right)\], we get,
\[ \Rightarrow \left( {y - \dfrac{3}{2}} \right) = \dfrac{1}{5}\left( {x - \dfrac{1}{2}} \right)\]
Taking LCM inside the brackets on both the sides, we get
\[ \Rightarrow \left( {\dfrac{{2y - 3}}{2}} \right) = \dfrac{1}{5}\left( {\dfrac{{2x - 1}}{2}} \right)\]
Cancelling out the denominator and solving further, we get,
\[ \Rightarrow 5\left( {2y - 3} \right) = \left( {2x - 1} \right)\]
\[ \Rightarrow 10y - 15 = 2x - 1\]
Solving further, we get,
\[ \Rightarrow 2x - 10y + 14 = 0\]
Dividing both sides by 2, we get
\[ \Rightarrow x - 5y + 7 = 0\]
Therefore, the equation of normal to the given curve is \[x - 5y + 7 = 0\]

Hence, the equations of the tangent and normal to the given curve are \[5x + y - 4 = 0\] and \[x - 5y + 7 = 0\] respectively.

Note:
In geometry, the tangent to a curve at a given point is a straight line that touches the curve at that specific point. The point where a tangent touches the curve is called the point of contact. Also, the normal to a curve is a line perpendicular to the tangent to the curve at that point. Hence, the normal and the tangent are perpendicular to each other. Now, we know that the slopes of perpendicular lines are negative reciprocals of one another, hence, the product of the slope of tangent to that of the normal is equal to \[ - 1\].