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  • Please help! - Limit , continuity and differentiability - JEE Main-18

Which one of the following points does not lie on the normal to the hyperbola, \frac{x^{2}}{16}-\frac{y^{2}}{9}=1         drawn at the point

(8,3\sqrt{3})\; ?

  • Option 1)

    (10,\: \sqrt{3})\;

  • Option 2)

    \left (13,-\frac{1}{\sqrt{3}} \right )

  • Option 3)

    \; \left (12, \;\frac{1}{\sqrt{3}} \right )

  • Option 4)

    \left ( 11,\sqrt{3} \right )

 

Answers (1)

best_answer

As we have learned

Equation of Normal -

Equation of normal to the curve  y = f(x) at the point  P(x1, y1) on the curve having a slope  MN  is 

(y-y_{1})=M_{N}(x-x_{1})


=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}(x-x_{1})

-

 

 we have 

\frac{x}{8}- \frac{24}{9}y' = 0 \\

y ' = \frac{9x}{16y} = \frac{9 \times 8}{16\times 3V_3 }= \frac{\sqrt 3}{2}

So equation of normal 

 (y- 3\sqrt 3 ) = \frac{-2 }{\sqrt 3 }(x-8 )\\ 2x + \sqrt 3 y - 25 = 0

 

 

 

 


Option 1)

(10,\: \sqrt{3})\;

Option 2)

\left (13,-\frac{1}{\sqrt{3}} \right )

Option 3)

\; \left (12, \;\frac{1}{\sqrt{3}} \right )

Option 4)

\left ( 11,\sqrt{3} \right )

Posted by

Himanshu

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