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

Hint:
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 \[\left( {\dfrac{4}{{{{( - 4)}^2}}},\dfrac{{2 \times 4}}{{ - 4}}} \right) = \left( {\dfrac{1}{4}, - 2} \right)\].