Question

How do you find the equations of both lines through point \[\left( {2, - 3} \right)\] that are tangent to the parabola \[y = {x^2} + x\]?

Answer 1

Hint: The derivative of any equation in the form \[y = f\left( x \right)\] gives the tangent of the function \[f\left( x \right)\] at point \[\left( {x,f\left( x \right)} \right)\]. Also the equation of a line that passes through point \[\left( {{x_1},{y_1}} \right)\] and have slope \[m\] is \[\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)\].

Complete step-by-step solution:
The given equation of a parabola is \[y = {x^2} + x\].
Differentiate the given equation with respect to \[x\] and obtain the tangent of the equation at \[\left( {x,f\left( x \right)} \right)\] as shown below.
\[y' = 2x + 1\]
\[ \Rightarrow m = 2x + 1\]
Where \[m\] represent the slope of a curve or tangent to the equation at point defined as \[\left( {x,f\left( x \right)} \right) = \left( {x,{x^2} + x} \right)\].
The slope of a line that passes through two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is calculated by the formula \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Use the slope formula, to find the slope of tangent that passes through the point \[\left( {2, - 3} \right)\] and the point \[\left( {x,{x^2} + x} \right)\] as follows:
\[\begin{array}{c}m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\\ = \dfrac{{\left( {{x^2} + x} \right) - \left( { - 3} \right)}}{{\left( x \right) - \left( 2 \right)}}\\ = \dfrac{{{x^2} + x + 3}}{{x - 2}}\end{array}\]
Put the above slope equivalent to the tangent \[2x + 1\] to the curve \[y = {x^2} + x\] and solve for \[x\] to obtain the point on the curve \[y = {x^2} + x\] from where the tangent passes through the point \[\left( {2, - 3} \right)\].
\[\dfrac{{{x^2} + x + 3}}{{x - 2}} = 2x + 1\]
\[ \Rightarrow {x^2} + x + 3 = \left( {2x + 1} \right)\left( {x - 2} \right)\]
\[ \Rightarrow {x^2} + x + 3 = 2{x^2} + x - 4x - 2\]
\[ \Rightarrow {x^2} - 4x - 5 = 0\]
Evaluate the quadratic equation as shown below.
\[ \Rightarrow {x^2} - 5x + x - 5 = 0\]
\[ \Rightarrow x\left( {x - 5} \right) + \left( {x - 5} \right) = 0\]
\[ \Rightarrow \left( {x - 5} \right)\left( {x + 1} \right) = 0\]
\[ \Rightarrow x = 5, - 1\]
Therefore, the slope of a curve at \[x = 5\] is calculated as,
\[\begin{array}{c}{m_1} = 2\left( 5 \right) + 1\\ = 11\end{array}\]
Similarly, the slope of a curve at \[x = - 1\] is calculated as,
\[\begin{array}{c}{m_2} = 2\left( { - 1} \right) + 1\\ = - 1\end{array}\]
Now obtain the equation of a line that passes through the point \[\left( {2, - 3} \right)\] and have a slope \[{m_1} = 11\].
\[ \Rightarrow y - \left( { - 3} \right) = 11\left( {x - 2} \right)\]
\[ \Rightarrow y + 3 = 11x - 22\]
\[ \Rightarrow y = 11x - 25\]
Similarly, obtain the equation of a line that passes through the point \[\left( {2, - 3} \right)\] and have a slope \[{m_2} = - 1\].
\[ \Rightarrow y - \left( { - 3} \right) = - 1\left( {x - 2} \right)\]
\[ \Rightarrow y + 3 = - x + 2\]
\[ \Rightarrow y = - x - 1\]
Thus, the equations of both lines through the point \[\left( {2, - 3} \right)\] that are tangent to the parabola \[y = {x^2} + x\] are \[y = 11x - 25\] and \[y = - x - 1\].

Note: Derivative of a linear equation is constant, It implies that slope of a line does not change with position of a point at which slope is calculated. Similarly, derivative of a curve or slope generally varies with position of a point on a curve at which slope is calculated.