Answer :
x^2 + y^2 = (3x^2 + 2y^2 - x)^2
2x + 2y f'(x) = 2(3x^2 + 2y^2 - x)(6x + 4y f'(x) - 1) = 36x^3 + 24x^2yf'(x) + 24xy^2 + 16y^3f'(x) - 4y^2 - 18x^2 - 8xyf'(x) + x
f'(x)(2y - 24x^2y - 16y^3 + 8xy) = 36x^3 + 24xy^2 - 4y^2 - 18x^2 - x
f'(x) = (36x^3 + 24xy^2 - 4y^2 - 18x^2 - x)/(2y - 24x^2y - 16y^3 + 8xy)
f'(0, 0.5) = -4(0.5)^2/(2(0.5) - 16(0.5)^3) = -1/(1 - 2) = -1/-1 = 1
Let the required equation be y = mx + c; where y = 0.5, m = 1, x = 0
0.5 = 1(0) + c = 0 + c
c = 0.5
Therefore, the tangent line at point (0, 0.5) is
y = x + 0.5
2x + 2y f'(x) = 2(3x^2 + 2y^2 - x)(6x + 4y f'(x) - 1) = 36x^3 + 24x^2yf'(x) + 24xy^2 + 16y^3f'(x) - 4y^2 - 18x^2 - 8xyf'(x) + x
f'(x)(2y - 24x^2y - 16y^3 + 8xy) = 36x^3 + 24xy^2 - 4y^2 - 18x^2 - x
f'(x) = (36x^3 + 24xy^2 - 4y^2 - 18x^2 - x)/(2y - 24x^2y - 16y^3 + 8xy)
f'(0, 0.5) = -4(0.5)^2/(2(0.5) - 16(0.5)^3) = -1/(1 - 2) = -1/-1 = 1
Let the required equation be y = mx + c; where y = 0.5, m = 1, x = 0
0.5 = 1(0) + c = 0 + c
c = 0.5
Therefore, the tangent line at point (0, 0.5) is
y = x + 0.5
Answer:
The equation of tangent is y=x+0.5
Step-by-step explanation:
Given: [tex]x^2+y^2=(3x^2+2y^2-x)^2[/tex]
We need to find the equation of tangent on curve at (0,0.5)
First we find slope of tangent using differentiation at given point.
[tex]x^2+y^2=(3x^2+2y^2-x)^2[/tex]
Differentiate both side w.r.t x
[tex]2x+2yy'=2(3x^2+2y^2-x)(6x+4yy'-1)[/tex]
Put x=0 and y=0.5 to solve for y'
[tex]2(0)+2(0.5)y'=2(3(0)^2+2(0.5)^2-0)(6(0)+4(0.5)y'-1)[/tex]
[tex]0+y'=2(0+0.5-0)(0+2y'-1)[/tex]
[tex]y'=2y'-1[/tex]
[tex]y'=1[/tex]
Derivative at given point is slope of tangent line.
Thus, Slope = 1 and point: (0,0.5)
Using point slope form to find the equation of tangent line.
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-0.5=1(x-0)[/tex]
[tex]y-0.5=x[/tex]
[tex]y=x+0.5[/tex]
Hence, The equation of tangent is y=x+0.5