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

JEE · Math · previous-year question

  1. A.$${3 \over {\sqrt 5 }}$$
  2. B.$${{\sqrt 15 } \over 2}$$
  3. C.$${{\sqrt 5 } \over 2}$$
  4. D.$${2 \over {\sqrt 5 }}$$correct

Answer

D. $${2 \over {\sqrt 5 }}$$

Explanation

Given line y = mx + 7$$\sqrt 3 $$ .....(1) Given hyperbola $${{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1$$ Here $${a^2} = 24$$ and $${b^2} = 18$$ We know the equation of normal to the hyperbola is $$y = mx \pm {{m\left( {{a^2} + {b^2}} \right)} \over {\sqrt {{a^2} - {b^2}{m^2}} }}$$ $$ \Rightarrow $$ $$y = mx \pm {{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}$$ .....(2) Comparing (1) and (2), we get $${{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}$$ = $$7\sqrt 3 $$ $$ \Rightarrow $$ 36m2 = 72 - 54m2 $$ \Rightarrow $$ 90m2 = 72 $$ \Rightarrow $$ m2 = $${{72} \over {90}}$$ $$ \Rightarrow $$ m = $${2 \over {\sqrt 5 }}$$

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