Question

Find the equation of the tangents to the curve $y = {x^3} + 2x - 4$, which are perpendicular to the line $x + 14y + 3 = 0$.

Answer 1

Hint: Here, we will proceed by equating the slopes of the required tangent to the given curve obtained through the formula i.e., the slope of the tangent to any curve $y = f\left( x \right)$ is given by $\dfrac{{dy}}{{dx}}$ to that obtained by using the concept that the slope of the tangent perpendicular to the given line having slope ${m_1}$ is given by ${m_2} = - \dfrac{1}{{{m_1}}}$.

Complete step-by-step answer:
Given equation of the curve is $y = {x^3} + 2x - 4{\text{ }} \to {\text{(1)}}$
Given equation of the line is $x + 14y + 3 = 0{\text{ }} \to {\text{(2)}}$
As we know that the general equation of any straight line having slope as m and y intercept as c is $y = mx + c{\text{ }} \to {\text{(3)}}$
By rearranging the given equation of line (i.e., equation (2)) in the same form as that of the general equation of line (i.e., equation (3)), we get
$\Rightarrow 14y = - x - 3 \\\ \Rightarrow y = \dfrac{{ - x - 3}}{{14}} \\\ \Rightarrow y = - \dfrac{x}{{14}} - \dfrac{3}{{14}} \\\ \Rightarrow y = \left( { - \dfrac{1}{{14}}} \right)x + \left( { - \dfrac{3}{{14}}{\text{ }}} \right) \to {\text{(4)}} \\\$
By comparing equations (3) and (4), we get
Slope of the given line is ${m_1} = - \dfrac{1}{{14}}$ and the y intercept of the given line is $c = - \dfrac{3}{{14}}{\text{ }}$
Also we know that the slope of any straight line perpendicular to the straight line having slope ${m_1}$ is given by ${m_2} = - \dfrac{1}{{{m_1}}}$
Given that the tangent to the given curve needs to be perpendicular to the given straight line.
The slope of the tangent to the given curve is ${m_2} = - \dfrac{1}{{{m_1}}} = - \dfrac{1}{{\left( { - \dfrac{1}{{14}}} \right)}} \\\ \Rightarrow {m_2} = 14{\text{ }} \to {\text{(5)}} \\\$
Differentiating the equation of the given curve (i.e., equation (1)) both sides with respect to x, we get

$$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^3} + 2x - 4} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^3}} \right) + \dfrac{d}{{dx}}\left( {2x} \right) + \dfrac{d}{{dx}}\left( { - 4} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} + 2 + 0 \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} + 2{\text{ }} \to {\text{(6)}} \\\$$

Since, the slope of the tangent to any curve $y = f\left( x \right)$ is given by $\dfrac{{dy}}{{dx}}$
So, ${m_2} = \dfrac{{dy}}{{dx}}$
By substituting the equations (5) and (6) in the above equation, we get

$$\Rightarrow 14 = 3{x^2} + 2 \\\ \Rightarrow 3{x^2} = 12 \\\ \Rightarrow {x^2} = 4 \\\ \Rightarrow x = \pm 2 \\\$$

By substituting x = 2 in equation (1), we get
$y = {2^3} + 2\left( 2 \right) - 4 = 8 + 4 - 4 \\\ \Rightarrow y = 8 \\\$
By substituting x = -2 in equation (1), we get
$y = {\left( { - 2} \right)^3} + 2\left( { - 2} \right) - 4 = - 8 - 4 - 4 \\\ \Rightarrow y = - 16 \\\$
When x = 2, we get y = 8 and when x = -2, we get y = -16
So, the tangents having slope of ${m_2} = 14$ to the given curve are drawn at points (2,8) and (-2,-16)
As, the equation of the straight line having slope m and passing through the point $\left( {{x_1},{y_1}} \right)$ is given by $y - {y_1} = m\left( {x - {x_1}} \right){\text{ }} \to {\text{(7)}}$
By using equation (7), the equation of the tangent having slope 14 and drawn from the point (2,8) to the given curve is
$y - 8 = 14\left( {x - 2} \right) \\\ \Rightarrow y = 14x - 28 + 8 \\\ \Rightarrow y = 14x - 20 \\\ \Rightarrow 14x - y - 20 = 0{\text{ }} \to {\text{(8)}} \\\$
By using equation (7), the equation of the tangent having slope 14 and drawn from the point (-2,-16) to the given curve is
$y - \left( { - 16} \right) = 14\left( {x - \left( { - 2} \right)} \right) \\\ \Rightarrow y + 16 = 14x + 28 \\\ \Rightarrow y = 14x + 12 \\\ \Rightarrow 14x - y + 12 = 0{\text{ }} \to {\text{(9)}} \\\$
Therefore, the required equations of tangents to the given curves are $14x - y - 20 = 0$ and $14x - y + 12 = 0$.

Note: In this particular problem, we have used the concept that the multiplication of the slopes of two straight lines which are perpendicular to each other are always equal to -1 i.e., ${m_1}{m_2} = - 1$. Here, the tangent to the given curve can be drawn through two points where the point in the first quadrant is (2,8) and that in the third quadrant is (-2,-16).