Find $m$ such that the line is normal to the given hyperbola
$begingroup$
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
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
$endgroup$
add a comment |
$begingroup$
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
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
$endgroup$
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
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
edited Jan 8 at 18:24
Maria Mazur
49.4k1360123
49.4k1360123
asked Jan 8 at 17:07
tatantatan
5,81462760
asked Jan 8 at 17:07
tatantatan
5,81462760
asked Jan 8 at 17:07
tatantatan
5,81462760
5,81462760
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $boxed{m= pm {2sqrt{3}over 3}}$
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $boxed{m= pm {2sqrt{3}over 3}}$
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
$endgroup$
add a comment |
$begingroup$
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 $boxed{m= pm {2sqrt{3}over 3}}$
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
$endgroup$
add a comment |
$begingroup$
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 $boxed{m= pm {2sqrt{3}over 3}}$
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
$endgroup$
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 $boxed{m= pm {2sqrt{3}over 3}}$
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
answered Jan 8 at 17:16
Maria MazurMaria Mazur
49.4k1360123
49.4k1360123
add a comment |
add a comment |
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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