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# Please help me with this calculus problem:?

1. Find the equation of tangent line of curve x(y^2)-2(x^2)y=6 at (-2,3)

2. Find the point at curve y=-1/3(x^2)+3 having the minimum distance to point (1,1) and compute the distanca

please someone help me as soon asa possible. i am going to have an examination, final exam on my university and i cant understand that...

T.T

i will not forget ur help

thank you very much

### 5 Answers

- sahsjingLv 71 decade agoFavorite Answer
1. Find the equation of tangent line of curve x(y^2)-2(x^2)y=6 at (-2,3)

Differentiate implicitly,

y^2 + x(2y)y' - 4xy - 2x^2y' = 0

Plug in given numbers: x = -2, y = 3

9 - 12y' + 24 - 8y' = 0

Solve for y',

y' = 33/20

Therefore, the equation of tangent line can be written as

y - 3 = (33/20)(x+2)

2. Find the point at curve y=-1/3(x^2)+3 having the minimum distance to point (1,1) and compute the distance.

Pick a point (x, y) from the curve. Let D be the distance from (1,1) to (x,y).

D^2 = (x-1)^2 + (y-1)^2 ......(1)

Differentiate (1) with respect to x,

2D dD/dx = 2(x-1) + 2(y-1)y' = 0 ......(2)

Simplify (2) and plug in y-1 = (-1/3)x^2+2 and y' = (-2/3)x,

x-1 + [(-1/3)x^2+2](-2/3)x = 0

Solve for x,

x = 1.9510

y = (-1/3)x^2 + 3 = 1.7312

D = √[(x-1)^2 + (y-1)^2] = 1.12 (the minimum distance)

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- NorthstarLv 71 decade ago
1. Find the equation of tangent line of curve x(y^2)-2(x^2)y=6 at (-2,3).

The tangent line will have the same slope as the curve at (-2,3). So find the derivative at the point (-2,3). Differentiate implicitly.

y² + 2xy(dy/dx) - 4xy - 2x²(dy/dx) = 0

(2xy - 2x²)(dy/dx) = -y² + 4xy

dy/dx = (-y² + 4xy)/(2xy - 2x²)

at (-2,3)

dy/dx = (-3² + 4(-2)(3))/(2(-2)(3) - 2(-2)²)

dy/dx = (-9 - 24)/(-12 - 8) = -33/-20 = 33/20

So the equation of the tangent line at (-2,3) is

y - 3 = (33/20)(x + 2)

y = (33/20)x + 33/10 + 3

y = (33/20)x + 63/10

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- PuggyLv 71 decade ago
1. x(y^2) - 2(x^2)y = 6 at (-2, 3)

First thing you do is differentiate implicitly, with respect to x.

(y^2) + x(2y)(dy/dx) - 2[2xy + (x^2)(dy/dx)] = 0

Simplifying even more, we get

(y^2) + 2xy(dy/dx) - 4xy - 2x^2(dy/dx) = 0

Move everything with a (dy/dx) to the right hand side,

2xy(dy/dx) - 2x^2(dy/dx) = -y^2 + 4xy

Factor (dy/dx) on the left hand side.

(dy/dx) [2xy - 2x^2] = -y^2 + 4xy

Now, isolate dy/dx by dividing 2xy - 2x^2 both sides.

(dy/dx) = [-y^2 + 4xy]/[2xy - 2x^2]

Let's factor the top and bottom to see if anything cancels. If not, the answer will be cleaner nonetheless anyway.

(dy/dx) = [-y(y - 4x)] / [2x(y - x)]

(dy/dx) = [y / (2x)] [4x - y]/[y - x]

In order to find the equation of the tangent line, we plug in x = -2 and y = 3 in the equation for dy/dx

m = [3/(2*(-2))] [4(-2) - 3]/[3 - (-2)]

m = [-3/4] [-11/5]

m = 33/20

Now that we have the slope, we can use the slope formula to get the equation of the line, letting (x1, y1) = (-2, 3), and

(x2, y2) = (x, y)

(y2 - y1) / (x2 - x1) = m

(y - 3) / (x - (-2)) = 33/20

20(y - 3) = 33(x + 2)

20(y - 3) = 33x + 66

y - 3 = (33/20)x + 33/10

y = (33/20)x + 33/10 + 3

y = (33/20)x + 63/10

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- Anonymous1 decade ago
#1

1.Take the derivative of the equation.

2. Use the slope formula, (y2-y1)/(x2-x1).

3. Set the slope of each tangent line equal to the derivative and solve for x.

Good luck!!

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- Anonymous1 decade ago
#1

y' = y^2 + x(2y)y' - 4xy-2(x)^2y' = 0

(y)y^2- 4xy = y' ( 2(x)^2 -x(2y) )

y' = [ (y)y^2- 4xy ] / [ 2(x)^2 -x(2y) ]

plug in the values and then you should get 33/20 which is your slope

your equation using point slope form should there for =

y-3 = 33/20 (x+2)

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