# Answer to Question #16308 in Calculus for hsd

Question #16308

Determine the equations of the tangent lines and of the normal lines to the curve

(x^2 + y^2)^(2/3) = xy

at the points (2, 2) and (-2,-2).

(x^2 + y^2)^(2/3) = xy

at the points (2, 2) and (-2,-2).

Expert's answer

(x^2+y^2)^2=x^3*y^3 => x^4+2x^2*y^2+y^4=(x*y)^3 => (find the derivate)

3x^3+2x*y^2+2*x^2*2*y*y'+4*y^3*y'=3*x^2*y^3+3*x^3*y^2*y' => y'(2,2)=-7/4,

y'(-2,-2)=7/4

Tangent: y-y0=y'(x-x0) => 1) y-2=(-7/4)*(x-2) =>

y=(-7x+22)/8; 2) y+2=(-7/4)*(x+2) => y=(-7x-22)/8

Normal:

y-y0=-(x-x0)/y' => 1) y=(4x+6)/7; 2) y=(4x-6)/7

3x^3+2x*y^2+2*x^2*2*y*y'+4*y^3*y'=3*x^2*y^3+3*x^3*y^2*y' => y'(2,2)=-7/4,

y'(-2,-2)=7/4

Tangent: y-y0=y'(x-x0) => 1) y-2=(-7/4)*(x-2) =>

y=(-7x+22)/8; 2) y+2=(-7/4)*(x+2) => y=(-7x-22)/8

Normal:

y-y0=-(x-x0)/y' => 1) y=(4x+6)/7; 2) y=(4x-6)/7

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