Consider the curve defined by the equation 4x^2+y^=7. Find the equation of the normal to the curve at the point (1,√3)

Question

 Has to show all steps of working, but I don't know how 

Answer 1

Since the point (1,√3) is on the graph, the equation must be 4x² + y² = 7.

 

Differentiating implicitly with respect to x, we have 8x + 2y(dy/dx) = 0

 

dy/dx = -4x/y

 

Slope of tangent at (1,√3) is -4(1)/√3 = -4/√3

 

Slope of normal = √3/4

 

Normal line has slope √3/4 and goes through the point (1,√3)

 

Equation of normal:  y - √3 = (√3/4)(x - 1)

 

Answer 2

your equation is defective, by omission of the exponent on y.  FIX IT! Thanks.