Question

If \[y = 4x + 3\] is parallel to the tangent of the parabola \[{y^2} = 12x\] , then its distance from the normal parallel to the given line is
\[1){\text{ }}\dfrac{{213}}{{\sqrt {17} }}\]
\[2){\text{ }}\dfrac{{219}}{{\sqrt {17} }}\]
\[3){\text{ }}\dfrac{{211}}{{\sqrt {17} }}\]
\[4){\text{ }}\dfrac{{210}}{{\sqrt {17} }}\]

Answer 1

Hint: We can find the slope of the tangent by differentiating the equation of parabola. Then find the slope of normal as the slope of the given equation is equal to the slope of normal because these lines are parallel to each other. Now because slope of perpendicular lines are negative reciprocal of one another, so find the value of y coordinate. Then put the y coordinate in the given equation of parabola by which you will get the value of x coordinate. Then find the distance from the point to the given equation of line. V

Complete step by step answer:
The general equation of a straight line is \[y = mx + c\] , ------- $(1)$
where m is the slope of the equation and c is the value where the line cuts the y-axis. In other words, c is called the intercept on the y-axis. The slope of any line perpendicular to a line with slope \[m\] is the negative reciprocal, that is \[ - \dfrac{1}{m}\] . Because the slopes of perpendicular lines are negative reciprocals of one another, therefore the slope of a normal is \[ - \dfrac{1}{m}\]
And the given equation of line in the question is \[y = 4x + 3\] --------- $(2)$
On comparing equations $(1)$ and $(2)$ we get , slope of the line \[y = 4x + 3\] , \[m = 4\]
It is given that the equation of the parabola is
\[{y^2} = 12x\] ---------- $(3)$
On differentiating with respect to \[x\] we get
\[2y\dfrac{{dy}}{{dx}} = 12\]
This gives us \[\dfrac{{dy}}{{dx}} = \dfrac{6}{y}\]
Therefore, the slope of a tangent on the parabola is \[\dfrac{{dy}}{{dx}} = \dfrac{6}{y}\]
As we know that normal line is perpendicular to the tangent and the slopes of two perpendicular lines are negative reciprocals of each other.
Hence slope of normal \[\left( {{m_N}} \right) = - \dfrac{y}{6}\]
Now in the question it is given that normal line and the given line of equation \[y = 4x + 3\] are parallel to each other . Also the slopes of parallel lines are equal. Therefore,
Slope of line having equation \[y = 4x + 3\] \[ = \] slope of normal line
\[ - \dfrac{y}{6} = 4\]
\[ \Rightarrow y = - 24\]
Put this value in equation $(3)$
\[\left( 3 \right) \Rightarrow {\text{ }}x = \dfrac{{24 \times 24}}{{12}}\]
\[ \Rightarrow {\text{ }}x = {\text{48}}\]
Therefore , point of normal becomes \[\left( {{\text{48, - 24}}} \right)\]
Formula to find the distance from the line \[\left( {ax + by + c = 0} \right)\] to a point \[\left( {{x_0},{y_0}} \right)\] is given by \[D{\text{ = }}\dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\]
So, to find the distance from \[\left( {{\text{48, - 24}}} \right)\] to line \[4x - y + 3 = 0\] ,
 \[a = 4\] , \[b = - 1\] , \[c = 0\] , \[{x_0} = 48\] and \[{y_0} = - 24\]
So, the distance from \[\left( {{\text{48, - 24}}} \right)\] to line \[4x - y + 3 = 0\] will be
\[D{\text{ = }}\dfrac{{\left| {4 \times 48 + \left( { - 1} \right) \times \left( { - 24} \right) + 3} \right|}}{{\sqrt {{{\left( 4 \right)}^2} + {{\left( { - 1} \right)}^2}} }}\]
\[D{\text{ = }}\dfrac{{\left| {192 + 24 + 3} \right|}}{{\sqrt {16 + 1} }}\]
\[D{\text{ = }}\dfrac{{\left| {219} \right|}}{{\sqrt {17} }}\]
Hence, the correct option is \[2){\text{ }}\dfrac{{219}}{{\sqrt {17} }}\]. So, option (B) is correct.

Note:
We can determine whether the lines are parallel or perpendicular by comparing their slopes. If the slopes are the same and the y-intercepts are different then the lines are parallel otherwise the lines are perpendicular. A tangent is a line that touches the parabola at exactly one point.