# Using implicit diffrentiation, determine the equation of the tangent at x=1 for the curve 2y-xy^2+x^2=1

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Determine the equation for the tangent at `x=1` for:

`2y-xy^2+x^2=1`

(1) Note that at `x=1` we have:

`2y-y^2+1=1`

`==>y^2-2y=0`

`==>y(y-2)=0`

`==>y=0,y=2`

Thus the points on the curve are (1,0) and (1,2)

(2) Using implicit differentiation we get:

`2y'-(y^2+2xyy')+2x=0`

** `d/(dx)2y=2(dy)/(dx)`

**`d/(dx)xy^2=d/(dx)[x]*y^2+x*d/(dx)y^2=y^2+x*2yy'`

**`d/(dx)x^2=2x,d/(dx)1=0`

Then:

`2y'-y^2-2xyy'+2x=0`

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

`y'=(y^2-2x)/(2-2xy)`

(3) Evaluating `y'` at the points (1,0) and (1,2) yield:

(a) at (1,0) `y'=(-2)/2=-1`

(b) at (1,2) `y'=(4-2)/(2-4)=-1` **By plugging in the values for x and y

Thus the slope of the tangent line at (1,0) is -1, and the slope of the tangent line at (1,2) is -1. We can use the point-slope form of a line to get the equation for the tangent lines:

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**The equation for the tangent lines when x=1 for `2y-xy^2+x^2=1` are:**

**At (1,0) `y=-x+1` and at (1,2) `y=-x+3` **

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The graph:

** The explicit functions can be found:

Write as a function of y, treating x as a constant:

`xy^2-2y+(1-x^2)=0` ; then use the quadratic formula:

`y=(2+-sqrt(4-4(x)(1-x^2)))/(2x)=(2+-sqrt(4(x^3-x+1)))/(2x)`

`=(1+-sqrt(x^3-x+1))/x`

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**Sources:**

At x=1;

2y-xy^2+x^2=1

2y+Y^2+1 =1

2y+y^2 = 0

y(y+2) = 0

y=0 OR y= -2

Then we will differnciate the both sides of the equation with respect to x.

2y'+x(2yy')+y^2+2x=0

Then y' = -(y^2+2x)/(2y+2)

When x=1 y has two answers as y=0 and y=-2

Let y=0

Then y'= -(0+2*1)/(2*0+2) = -1

So the gradient of the tangent is -1.

Then we can write the equation of tangent as (y-y1)=m(x-x1)

Here y1=0 and x1=1 and m=-1

Therefore (y-0) = -1(x-1)

**So the equation of tangent is y= -x+1**

Let y=-2

Then y'= -((-2)^2+2*1)/(2*(-2)+2) = -1

So the gradient of the tangent is -1.

Then we can write the equation of tangent as (y-y1)=m(x-x1)

Here y1=-2 and x1=1 and m=-1

Therefore (y-2) = -1(x-1)

**So the equation of tangent is y= -x+1+2 =-x+3**

**So finally there are two equations for tangent at x=1**

** y= -x+1**

**y= -x+1+2 =-x+3**