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

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.