Q. FIND THE EQUATION OF THE NORMAL TO THE CURVE- x3 + y3 = 8xy AT THE POINT WHERE IT MEETS THE CURVE y2 = 4x . OTHER THAN THE ORIGIN.

Ans:Hello Student,Please find answer to your question belowLets find the point of intersection:…......(1)Put in (1)Point of intersection other than origin(4, 4)Lets find the slope of normal:Differentiate (1)Slope: 1Equation of normal:

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Grade Select GradeDifferential Calculus

Q. FIND THE EQUATION OF THE NORMAL TO THE CURVE-
x³ + y³ = 8xy
AT THE POINT WHERE IT MEETS THE CURVE y² = 4x . OTHER THAN THE ORIGIN.

Bharat Makkar

11 Years agoGrade Select Grade

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

Lets find the point of intersection:
$x^3 + y^3 = 8xy$ …......(1)
$x^3 = y(8x-y^2)$
$y^2 = 4x$
$x^3 = y(8x-4x)$
$x^3 = 4xy$
$x^2 = 4y$
Put in (1)
$x^3 + (\frac{x^2}{4})^3 = 8x.\frac{x^2}{4}$
$x^3 + \frac{x^6}{64} = 2x^3$
$\frac{x^6}{64} = x^3$
$x^3(x^3-64) = 0$
$x = 0, 4$
Point of intersection other than origin
(4, 4)
Lets find the slope of normal:
$\frac{-1}{y'}$
Differentiate (1)
$3x^2 + 3y^2y' = 8y+8xy'$
$3x^2 - 8y = (8x-3y^2)y'$
$3.4^2 - 8.4 = (8.4-3.4^2)y'$
$16 = (-16)y'$
$y' = -1$
Slope: 1
Equation of normal:
$(y-4)=(x-4)$
$y=x$