The two curves given represented by the equations 5y-2x+y^3 -x^2y = 0 and 5x+2y+x^4-x^3y^2=0.
At the origin x = 0 and y = 0
Substituting these values in the equation of the curve 5y-2x+y^3 -x^2y = 0 satisfies the equation as both the sides are equal to 0. The same is the case with the curve 5x+2y+x^4-x^3y^2=0.
To show that the tangent to the curves at the point (0, 0) is perpendicular, the value of `dy/dx ` for both the curves is required. The slope of the tangent at (0,0) is given by the value of `dy/dx` at that point.
Using implicit differentiation, 5y-2x+y^3 -x^2y = 0 gives
`5*dy/dx - 2 + 3x^2(dy/dx) - 2x*y - x^2*(dy/dx) = 0`
=> `(dy/dx)(5 + 3x^2 - x^2) = 2xy + 2`
=> `(dy/dx) = (2xy + 2)/(5 + 3x^2 - x^2)`
At the origin
`dy/dx = (0 + 2)/(5 + 0 - 0) = 2/5`
For the curve 5x+2y+x^4-x^3y^2=0
`5 + 2*(dy/dx) + 4x^3 - 3x^2*y^2 - x^2*2y*(dy/dx) = 0`
=> `(dy/dx)(2 - x^2*2y) = 3x^2*y^2 - 4x^3 - 5`
=> `dy/dx = (3x^2*y^2 - 4x^3 - 5)/(2 - x^2*2y)`
At the origin `dy/dx = -5/2`
It is seen that the product of the slope of the two tangents is `(2/5)*(-5/2) = -1` .
This proves that for the curves 5y-2x+y^3 -x^2y = 0 and 5x+2y+x^4-x^3y^2=0 the tangents at the slope are perpendicular to each other.