Question

If the line $y=mx+7\sqrt{3}$ is normal to the hyperbola $\dfrac{{{x}^{2}}}{24}-\dfrac{{{y}^{2}}}{18}=1$ , then a value of $m$ is?
A. $\dfrac{\sqrt{5}}{2}$
B. $\dfrac{3}{\sqrt{5}}$
C. $\dfrac{2}{\sqrt{5}}$
D. $\dfrac{\sqrt{15}}{2}$

Answer 1

Hint: We will start with finding the standard equation of the normal to the hyperbola, $y=mx\pm \dfrac{m\left( {{a}^{2}}+{{b}^{2}} \right)}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}}$ and then we will move on to compare it with the equation of hyperbola, $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Then, we will relate these with hyperbola given in the question. Having found that, we will compare it with the equation of normal that is given in the question, then we will get the value of m.

Complete step-by-step answer:
We know that the equation of the normal of slope of $m$ to the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ , are given by $y=mx\pm \dfrac{m\left( {{a}^{2}}+{{b}^{2}} \right)}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}}\text{ }.......\text{Equation 1}\text{.}$ .
Now to find the equation of normal to the hyperbola $\dfrac{{{x}^{2}}}{24}-\dfrac{{{y}^{2}}}{18}=1$ we will apply the formula which we wrote down in equation 1, that is $y=mx\pm \dfrac{m\left( {{a}^{2}}+{{b}^{2}} \right)}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}}$
Now we have with us the equation of hyperbola given in the question as: $\dfrac{{{x}^{2}}}{24}-\dfrac{{{y}^{2}}}{18}=1$ , comparing it with the standard equation of hyperbola: $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ , We get the value of ${{a}^{2}}$ and ${{b}^{2}}$
Therefore, ${{a}^{2}}=24$ and ${{b}^{2}}=18$ ,
We will put these values in equation of the normal:
  $\begin{align}
  & \Rightarrow y=mx\pm \dfrac{m\left( {{a}^{2}}+{{b}^{2}} \right)}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}} \\
 & \Rightarrow y=mx\pm \dfrac{m\left( 24+18 \right)}{\sqrt{24-18{{m}^{2}}}} \\
 & \Rightarrow y=mx\pm \dfrac{m\left( 42 \right)}{\sqrt{24-18{{m}^{2}}}}\text{ }........\text{ Equation 2}\text{.} \\
\end{align}$
We are given the equation of normal in the question as: $y=mx+7\sqrt{3}$ , on comparing it with Equation 2, that is $y=mx\pm \dfrac{m\left( 42 \right)}{\sqrt{24-18{{m}^{2}}}}$
We have: $\pm \dfrac{m\left( 42 \right)}{\sqrt{24-18{{m}^{2}}}}=7\sqrt{3}$ now we will solve this equation:
$\begin{align}
  & \Rightarrow \pm \dfrac{m\left( 42 \right)}{\sqrt{24-18{{m}^{2}}}}=7\sqrt{3} \\
 & \Rightarrow \pm \dfrac{m\left( 7\times 6 \right)}{\sqrt{24-18{{m}^{2}}}}=7\sqrt{3}\Rightarrow \pm \dfrac{m\left( 6 \right)}{\sqrt{24-18{{m}^{2}}}}=\sqrt{3} \\
\end{align}$
Now squaring both the sides: we will get
$\begin{align}
  & \Rightarrow \dfrac{36{{m}^{2}}}{24-18{{m}^{2}}}=3\Rightarrow \dfrac{12{{m}^{2}}}{24-18{{m}^{2}}}=1 \\
 & \Rightarrow 12{{m}^{2}}=24-18{{m}^{2}} \\
 & \Rightarrow 30{{m}^{2}}=24\Rightarrow {{m}^{2}}=\dfrac{24}{30} \\
 & \Rightarrow {{m}^{2}}=\dfrac{4}{5}\Rightarrow m=\sqrt{\left( \dfrac{4}{5} \right)}\Rightarrow m=\pm \dfrac{2}{\sqrt{5}} \\
\end{align}$
Therefore the correct option is C.

Note: In this question, we have been given the standard form of hyperbola, but if it was not given, then we must convert to standard form and then only we can compare and get the value of a and b accurately. After getting $\pm \dfrac{m\left( 42 \right)}{\sqrt{24-18{{m}^{2}}}}=7\sqrt{3}$ equation, we must solve carefully, trying to simplify and then take square on both sides. If a student does not simplify terms, then calculations might get tougher.