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Find the equation of normal to the curve X^3+Y^3=8XY at the point where it is meet by the curve Y^2=4X, other than origin??Ans =Y=X.Plz give the full solution.

Find the equation of normal to the curve X^3+Y^3=8XY at the point where it is meet by the curve Y^2=4X, other than origin??Ans =Y=X.Plz give the full solution.

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2 Answers

Arun
25750 Points
6 years ago
Point of intersection other than origin(4, 4)Lets find the slope of normal:\frac{-1}{y`}Differentiate (1)3x^2 + 3y^2y` = 8y+8xy`3x^2 - 8y = (8x-3y^2)y`3.4^2 - 8.4 = (8.4-3.4^2)y`16 = (-16)y`y` = -1Slope: 1Equation of normal:(y-4)=(x-4)y=xLets find the point of intersection:x^3 + y^3 = 8xy…......(1)x^3 = y(8x-y^2)y^2 = 4xx^3 = y(8x-4x)x^3 = 4xyx^2 = 4yPut in (1)x^3 + (\frac{x^2}{4})^3 = 8x.\frac{x^2}{4}x^3 + \frac{x^6}{64} = 2x^3\frac{x^6}{64} = x^3x^3(x^3-64) = 0x = 0, 4Please find answer to your question below
Arun
25750 Points
6 years ago
 
The meeting point of above two curve are (0,0) and (4,4)
So we have to find out normal to curve x3 + y3 = 8xy at point (4,4)
 differentiating above equation
3x2 +3y2m = 8y +8xm    where m = y’
putting x =4 and y=4
48 + 48m = 32 + 32m
16m = -16
m = -1
slope of normal at (4,4) = -1/m = 1
So equation of normal
(y-4) = 1(x-4)
So y = x.
 
 
Regards
Arun (askIITians forum expert)

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