Question

find the area bounded by the parabola (y-2)^2=x-1, the tangent at the point (2,3) and the x-axis .

Answer 1

pls integrate wrt y so that i can check it with my answer...

Answer 2

@ganeshie8 @freckles @UnkleRhaukus

Answer 3

Find area under parabola from x =0 to x=1

Find eqaution of the tangent

Find the area under tangent from its x intersept to x=1

Subtract area under tangent from area under parabola



Area under parabola

y = 4x^2

This parabola passes through (0,0)

Definite Integrate from x =0 to point of tangent x= 1

Integral y = (4x^3)/3

Definite Integrate from x =0 to point of tangent x= 1 = 4 x 13/3 – 0 = 4/3 units



Equation of tangent line

Slope of parabola at (1,4)

Differentiate y = 4x^2

dy/dx = 8x = slope

slope at x=1 would be 8 x1 = 8

equation of tangent

y = mx + b

y = 8x + b

since this line passes through (1,4)

4 = 8 x 1 + b

b = -4

tangent equation y= 8x – 4

x intercept at y=0

0= 8x -4

X=1/2



Area under tangent from x= ½ to x=1

Integrate y = 8x - 4

Integral y = (8x^2)/2 – 4x

Definite integral from x= ½ to x= 1

((8 x 1)/2 – 4 x 1) – ((8 x ½ 2/2) – 4x1/2)

0 – (-1) = 1 unit



Subtract area under tangent from area under parabola



4/3 – 1 = 1/3 units – Answer :)

Answer 4

this question is unrelated to my question.. @Thehulk49 ??

Answer 5

I like to graph my equations first:
\[y=\frac{x}{2}+2 \\ (y-2)^2=x-1 \\ \\ y=0 \\ .... \\ \text{ solving for intersections } \\ (\frac{x}{2}-2)^2=x-1 \\ \frac{x^2}{4}-2x+4=x-1 \\ x^2-8x+16=4x-4 \\ x^2-12x+20=0 \\ (x-10)(x-2)=0 \\ x=10, 2 \\ \]



if I didn't make a mistake in finding the tangnet line or finding the intersections