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Differential Calculus

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19 Mar 2007 15:27:03 IST

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AOD!!!!!!! plz help

None

Q->1) find the condition that the line xcos

+ysin

= P may touch the curve

(x/a)^m + (y/b)^m = 1

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Comments (8)

Ashish Kumar

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19 Mar 2007 17:32:50 IST

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is this a question of calculus?
it seems like a conics question man

Himanshu

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19 Mar 2007 17:33:59 IST

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nope its a qs of calculas only

Reshma Sajan

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19 Mar 2007 18:58:47 IST

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is a constant.. then this mite work :

cot

= - (m/n) (x^m-1/y^n-1) (bⁿ/a^m)

{ Method:

Slope of curve = Slope of line or tangent.

Differentiate curve and find dy/dx
Diff. line and find dy/dx
equate both. }

maybe it mite be wrong... I'm new to all this stuff

Himanshu

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19 Mar 2007 19:00:50 IST

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i m rating u for ur try bt i m not satisfied

neways thanx for trying

Himanshu

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20 Mar 2007 09:35:19 IST

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anyone

plzzzzzzzzzzzzz heeeeeeeeeelllllllllllllllppppppppppppp!!!!!!!!!!!!!

Divya Alok Gupta

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20 Mar 2007 10:50:47 IST

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(x/a)^m + (y/b)^m = 1 is the curve and we have to prove that xcos

+ y sin

= P is tangential to it as in find the condition on

then y = (P - xcos

) cosec

(x/a)^m + (P-xcos

/bsin

)^m = 1

Now u put the condition that since the line touches the curve only at one point the polynomial has one solution repeated or many imaginary solutions and 2 repeated roots depending on m and solve the polynomial

(xbsin

/ a)^m + (P-xcos

)^m = (bsin

)^m

Now i guess you'll have problems solving it but its worth a try i have done the most logical part u have to just do manual solving!!!!

Plz do rate me for my small but sincere effort!!!

Jyotibrata Bhattacharjee

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20 Mar 2007 12:31:24 IST

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(Q) must be const

To find the cond.

diff 1

we get dy/dx= -cotQ

Diff 2

we get dy/dx=-(x/y)^(m-1) (b/a)^m

So, as the slope must be equal,

cotQ=(x/y)^m-1 (b/a)^m

Rate me plz

Ayshwar Pandey

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20 Mar 2007 12:58:50 IST

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hi himanshu,
clearly, dy/dx of line=-cot(theta)
dy/dx of curve=-(b/a)*m. (x/y)*m-1

Equating both,
we get the condition as
cot(theta)=(b/a)*m. (x/y)*(m-1)

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