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?

• A. $\frac{213}{\sqrt{17}}$
• B. $\frac{219}{\sqrt{17}}$
• C. $\frac{211}{\sqrt{17}}$
• D. $\frac{210}{\sqrt{17}}$

Answer 1

The correct option is B

$\frac{219}{\sqrt{17}}$

Explanation for the correct option:

Step 1: Finding the slope of line.

The equation of a line given is $y=4x+3$ compare it with standard equation of line $y=mx+c$

Then the slope of line is $m=4$

Step 2: Find the tangent of parabola

Now, get the tangent to the parabola $y^2=12x$ by differentiating it with respect to $x$

$$\begin{gathered} 2y\left(\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.\right)=12 \\ \Rightarrow y\left(\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.\right)=6 \\ \Rightarrow \left(\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.\right)=\frac{6}{y} \end{gathered}$$

Step 3: Find the coordinate point

The slope of the normal =$\frac{-1}{{\displaystyle \frac{dy}{dx}}}$

$\begin{array}{rcl}4& =& -1/(6/y)\\ & \Rightarrow & 4=-y/6\\ & \Rightarrow & y=-24\end{array}$

Substitute the value of $y$ in $y^2=12x$to get $x$ coordinate.

$$\begin{gathered} y^2=12x \\ \Rightarrow 24\times 24=12\times x \\ \Rightarrow x=48 \end{gathered}$$

So, the coordinate point $(x_1,y_1)=(48,-24)$

Thus, the line $ax+by=c$. Therefore comparing it with given equation $y=4x+3$ we get $(a,b)=(4,1)$

Step 4: Find the distance ($d$)

Then the distance $d$ from the tangent point $(x_1,y_1)=(48,-24)$ to the parallel line $y=4x+3$

$\begin{array}{rcl}d& =& \raisebox{1ex}{$\left[a{x}_1+b{y}_1–c\right]$}\!\left/ \!\raisebox{-1ex}{$\sqrt{a^2+b^2}$}\right.\\ & =& \raisebox{1ex}{$\left(4\times 48-1\times \left(-24\right)+3\right)$}\!\left/ \!\raisebox{-1ex}{$\sqrt{\left(16+1\right)}$}\right.\\ & =& \raisebox{1ex}{$219$}\!\left/ \!\raisebox{-1ex}{$\sqrt{17}$}\right.\end{array}$

Hence, the correct option is (B).