find an equation of the line L which is perpendicular to the tangent line

Question

to the graph of y^2(x^2+y^2)=2x^2 at(1,1), and goes through the same point

Answer 1

The equation of the curve is y²(x² + y²) = 2x²

Differentiating  on each side with respect of x .

d/dx(y²(x² + y²)) =d/dx(2x²)

Using product rule differentiation.

y² d/dx (x² + y²) + (x² + y²) d/dx (y²) = 4x

y² (2x + 2ydy/dx) +  (x² + y²) 2y dy/dx = 4x

2xy² + 2y (1 +(x² + y²) ) dy/dx =4x

dy/dx = ( 4x - 2xy² )/ 2y (1 +(x² + y²) )

Substitute the values (x , y ) = (1,1) in above equation.

dy/dx = ( 4(1) - 2(1)(1)² )/ 2(1) (1 +(1² + 1²) ) = 2 / 6 =1 / 3

This is the slope of tangent line to the curve at (1, 1).

m = 1 / 3

Slope of the perpendicular line to the tangent line is -1/m = -3

Find an equation of the line L which is perpendicular to the tangent line, substitute the values of m = -3 and (x, y ) = (1, 1).  in the slope intercept form of an equation.

1 = -3 (1) + b

b = 4

Substitute m = -3 and b = 4 in y = mx + b.

y = -3x + 4

Equation of the line L which is perpendicular to the tangent line is  y = -3x + 4.