Question

The equation of the tangent to the parabola ${y^2} = 16x$ and perpendicular to the line $x - 4y - 7 = 0$ is
A) $4x + y + 1 = 0$
B) $4x + y + 7 = 0$
C) $4x + y - 1 = 0$
D) $4x + y - 7 = 0$

Answer 1

Here, we will find the equation of the line which is perpendicular to the given line by using the slope-intercept form. For this, we will use the condition of perpendicularity. Then, we will use the condition of tangency to find the tangent to the given parabola.

Formula used:
We will use the following formulas:

1. If ${m_1}$ and ${m_2}$ are the slopes of two perpendicular lines, then ${m_1} \times {m_2} = - 1$.
2. Condition of tangency to parabola: $c = \dfrac{a}{m}$

Complete step by step solution:
The given equation of parabola is ${y^2} = 16x$. This parabola will open in the rightward direction.
Let us compare the equation of the given parabola to the general form of rightward-open parabola i.e., ${y^2} = 4ax$. We have,
$\begin{array}{l}{y^2} = 16x = 4ax\\\ \Rightarrow a = 4\end{array}$
The given equation of line is $x - 4y - 7 = 0$.
We will rewrite this line in the slope-intercept form i.e., we will write it in the form $y = mx + c$. We get,
$4y = x - 7$
Dividing both side by 4, we get
$\Rightarrow y = \dfrac{1}{4}x - \dfrac{7}{4}$………………………………$\left( 1 \right)$
Let the slope and $y -$ intercept of the line in equation $\left( 1 \right)$ be denoted as ${m_1}$ and ${c_1}$ respectively. Therefore, from equation $\left( 1 \right)$, we have
${m_1} = \dfrac{1}{4}$
${c_1} = - \dfrac{7}{4}$
We have to find the equation of a line which is tangent to the parabola ${y^2} = 16x$ and perpendicular to the line $x - 4y - 7 = 0$.
Let the required line be $y = {m_2}x + {c_2}$ where ${m_2}$ and ${c_2}$ are its slope and $y -$ intercept respectively.
We know that if ${m_1}$ and ${m_2}$ are the slopes of two perpendicular lines, then ${m_1} \times {m_2} = - 1$.
Substituting ${m_1} = \dfrac{1}{4}$ in the formula ${m_1} \times {m_2} = - 1$, we get
$\begin{array}{l}\dfrac{1}{4} \times {m_2} = - 1\\\ \Rightarrow {m_2} = - 4\end{array}$
Now, the equation of the required line becomes $y = - 4x + {c_2}$.
We have to find the value of ${c_2}$.
We will use the condition for a line $y = mx + c$ to be a tangent to a parabola.
Substituting ${m_2} = - 4$ and $a = 4$ in the formula $c = \dfrac{a}{m}$, we get
${c_2} = \dfrac{a}{{{m_2}}} = \dfrac{4}{{ - 4}}$
Simplifying the expression, we get
$\Rightarrow {c_2} = - 1$
Thus, the equation of the required line becomes $y = - 4x - 1$.
Therefore, the equation of the line which is tangent to the parabola ${y^2} = 16x$ and perpendicular to the line $x - 4y - 7 = 0$ is $4x + y + 1 = 0$.

So, the correct option is option A.

Note:
Tangent to a parabola is a line which touches the parabola at one point. A parabola is a type of conic section which is open from one side. The point of contact of a tangent $y = mx + \dfrac{a}{m}$ with the parabola ${y^2} = 4ax$ is $\left( {\dfrac{a}{{{m^2}}},\dfrac{{2a}}{m}} \right)$.
In the given problem, the tangent to the parabola ${y^2} = 16x$ is $4x + y + 1 = 0$.
So, the point of contact becomes $\left( {\dfrac{4}{{{{( - 4)}^2}}},\dfrac{{2 \times 4}}{{ - 4}}} \right) = \left( {\dfrac{1}{4}, - 2} \right)$.