Find $m$ such that the line is normal to the given hyperbola












0












$begingroup$



Find $m$ such that $y=mx+frac{25}{sqrt3}$ is normal to $$frac{x^2} {16}-frac{y^2}9=1$$




How to go about this question. I don't find any clue.






calculus geometry conic-sections coordinate-systems







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  • $begingroup$
    Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
    $endgroup$
    – EuxhenH
    Jan 8 at 17:14












  • $begingroup$
    @EuxhenH I get $dy/dx=9x/16y$. Now?
    $endgroup$
    – tatan
    Jan 8 at 17:16










  • $begingroup$
    askiitians.com/iit-jee-coordinate-geometry/…
    $endgroup$
    – lab bhattacharjee
    Jan 8 at 17:25
















0












$begingroup$



Find $m$ such that $y=mx+frac{25}{sqrt3}$ is normal to $$frac{x^2} {16}-frac{y^2}9=1$$




How to go about this question. I don't find any clue.






calculus geometry conic-sections coordinate-systems







share|cite|improve this question











$endgroup$












  • $begingroup$
    Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
    $endgroup$
    – EuxhenH
    Jan 8 at 17:14












  • $begingroup$
    @EuxhenH I get $dy/dx=9x/16y$. Now?
    $endgroup$
    – tatan
    Jan 8 at 17:16










  • $begingroup$
    askiitians.com/iit-jee-coordinate-geometry/…
    $endgroup$
    – lab bhattacharjee
    Jan 8 at 17:25














0












0








0





$begingroup$



Find $m$ such that $y=mx+frac{25}{sqrt3}$ is normal to $$frac{x^2} {16}-frac{y^2}9=1$$




How to go about this question. I don't find any clue.






calculus geometry conic-sections coordinate-systems







share|cite|improve this question











$endgroup$





Find $m$ such that $y=mx+frac{25}{sqrt3}$ is normal to $$frac{x^2} {16}-frac{y^2}9=1$$




How to go about this question. I don't find any clue.






calculus geometry conic-sections coordinate-systems




calculus geometry conic-sections coordinate-systems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 18:24









greedoid

39.7k114798




39.7k114798










asked Jan 8 at 17:07









tatantatan

5,63362655




5,63362655












  • $begingroup$
    Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
    $endgroup$
    – EuxhenH
    Jan 8 at 17:14












  • $begingroup$
    @EuxhenH I get $dy/dx=9x/16y$. Now?
    $endgroup$
    – tatan
    Jan 8 at 17:16










  • $begingroup$
    askiitians.com/iit-jee-coordinate-geometry/…
    $endgroup$
    – lab bhattacharjee
    Jan 8 at 17:25


















  • $begingroup$
    Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
    $endgroup$
    – EuxhenH
    Jan 8 at 17:14












  • $begingroup$
    @EuxhenH I get $dy/dx=9x/16y$. Now?
    $endgroup$
    – tatan
    Jan 8 at 17:16










  • $begingroup$
    askiitians.com/iit-jee-coordinate-geometry/…
    $endgroup$
    – lab bhattacharjee
    Jan 8 at 17:25
















$begingroup$
Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
$endgroup$
– EuxhenH
Jan 8 at 17:14






$begingroup$
Start by finding the slope of the tangent to the curve at any point. Hint: Implicit differentiation.
$endgroup$
– EuxhenH
Jan 8 at 17:14














$begingroup$
@EuxhenH I get $dy/dx=9x/16y$. Now?
$endgroup$
– tatan
Jan 8 at 17:16




$begingroup$
@EuxhenH I get $dy/dx=9x/16y$. Now?
$endgroup$
– tatan
Jan 8 at 17:16












$begingroup$
askiitians.com/iit-jee-coordinate-geometry/…
$endgroup$
– lab bhattacharjee
Jan 8 at 17:25




$begingroup$
askiitians.com/iit-jee-coordinate-geometry/…
$endgroup$
– lab bhattacharjee
Jan 8 at 17:25










1 Answer
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$begingroup$

Say it is normal at point $T(a,b)inmathcal{H}$, then from the derivative we get $$ {aover 8} - {2bkover 9}=0$$ where $$k = y'(a) = -{1over m}$$



So we have $$k={9aover 16b}implies m = -{16bover 9a}$$



Since $T$ is on this normal we have $$b = -{16bover 9a} cdot a +{25over sqrt{3}}implies b= 3sqrt{3}$$



Plugin in $${a^2over 16}-{27over 9} =1implies a=pm 8$$



So $boxed{m= pm {2sqrt{3}over 3}}$






share|cite|improve this answer









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    1 Answer
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    active

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    $begingroup$

    Say it is normal at point $T(a,b)inmathcal{H}$, then from the derivative we get $$ {aover 8} - {2bkover 9}=0$$ where $$k = y'(a) = -{1over m}$$



    So we have $$k={9aover 16b}implies m = -{16bover 9a}$$



    Since $T$ is on this normal we have $$b = -{16bover 9a} cdot a +{25over sqrt{3}}implies b= 3sqrt{3}$$



    Plugin in $${a^2over 16}-{27over 9} =1implies a=pm 8$$



    So $boxed{m= pm {2sqrt{3}over 3}}$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Say it is normal at point $T(a,b)inmathcal{H}$, then from the derivative we get $$ {aover 8} - {2bkover 9}=0$$ where $$k = y'(a) = -{1over m}$$



      So we have $$k={9aover 16b}implies m = -{16bover 9a}$$



      Since $T$ is on this normal we have $$b = -{16bover 9a} cdot a +{25over sqrt{3}}implies b= 3sqrt{3}$$



      Plugin in $${a^2over 16}-{27over 9} =1implies a=pm 8$$



      So $boxed{m= pm {2sqrt{3}over 3}}$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Say it is normal at point $T(a,b)inmathcal{H}$, then from the derivative we get $$ {aover 8} - {2bkover 9}=0$$ where $$k = y'(a) = -{1over m}$$



        So we have $$k={9aover 16b}implies m = -{16bover 9a}$$



        Since $T$ is on this normal we have $$b = -{16bover 9a} cdot a +{25over sqrt{3}}implies b= 3sqrt{3}$$



        Plugin in $${a^2over 16}-{27over 9} =1implies a=pm 8$$



        So $boxed{m= pm {2sqrt{3}over 3}}$






        share|cite|improve this answer









        $endgroup$



        Say it is normal at point $T(a,b)inmathcal{H}$, then from the derivative we get $$ {aover 8} - {2bkover 9}=0$$ where $$k = y'(a) = -{1over m}$$



        So we have $$k={9aover 16b}implies m = -{16bover 9a}$$



        Since $T$ is on this normal we have $$b = -{16bover 9a} cdot a +{25over sqrt{3}}implies b= 3sqrt{3}$$



        Plugin in $${a^2over 16}-{27over 9} =1implies a=pm 8$$



        So $boxed{m= pm {2sqrt{3}over 3}}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 17:16









        greedoidgreedoid

        39.7k114798




        39.7k114798






























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