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# Engineering

If the line y=mx+7\sqrt{3} is normal to the hyperbola \frac{x^{2}}{24}-\frac{y^{2}}{18}=1   , then a value of m is :

  • Option 1)

    \frac{\sqrt{5}}{2}          

  • Option 2)

    \frac{\sqrt{15}}{2}

  • Option 3)

    \frac{2}{\sqrt{5}}

  • Option 4)

    \frac{3}{\sqrt{5}}

 

Answers (1)
V Vakul

     y=mx+7\sqrt{3}

given hyperbola      \frac{x^{2}}{24}-\frac{y^{2}}{18}=1

Normal to hyperbola is slope form

        y=mx\mp \frac{m\left ( a^{2}+b^{2} \right )}{\sqrt{a^{2}-m^{2}b^{2}}}

  compare this

     7\sqrt{3}=\frac{m\left ( 24 +18\right )}{\sqrt{24-18m^{2}}}

        m=\frac{2}{\sqrt{5}}

      


Option 1)

\frac{\sqrt{5}}{2}          

Option 2)

\frac{\sqrt{15}}{2}

Option 3)

\frac{2}{\sqrt{5}}

Option 4)

\frac{3}{\sqrt{5}}

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