Question

Find the radius of curvature at origin for the curves
(ii)
co
x3 – 2x²y + 3xy2 – 4y2 + 5x2 - 6xy+7y2–8y = 0
xºy-xy+ 2x²y-2xy2 + 2y2 – 3x2 + 3xy-4x = 0
in 2x + 4x+y + xy2 + 5yº + x2–2xy + y2–4x = 0
iv) x + y2 – 2x2 + 6y = 0.
v) 5x3 + 7y3 + 4x²y + 2x2 + 3xy + y2 + 4x = 0.
fthe equiang​

Answer 1

huhsush since jwid is a right triangle hint of happy anniversary both of us and the paradice out of the day Happy birthday Bhaiya Bhabhi and Ashish ji long

Answer 2

4/5

Step-by-step explanation:

f=x3 – 2x²y + 3xy2 – 4y2 + 5x2 - 6xy+7y2–8y

f(x)=0

f(y)=-8

f(xx)=10

f(yy)=14

f(xy)=-6

apply the formula

radius of curvature=(fx^2+fy^2)^3/2÷fxxfy^2-2fxyfxfy+fyyfx^2

therefore , we get,

radius of curvature=4/5