# Given the equation y^2-x+2y-3=0determine dy/dx, the equation of the line tangent to the graphof the equation at point (0, 3), and the line y=1/4 (x)=3 is tangent to the graph at point p determine...

Given the equation y^2-x+2y-3=0

determine dy/dx, the equation of the line tangent to the graphof the equation at point (0, 3), and the line y=1/4 (x)=3 is tangent to the graph at point p determine the coordinates of point P.

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`y^2-x+2y -3 = 0`

Implicit differentiation,

`2y(dy) -(dx)+2(dy) -0 = 0`

`(2y+2)(dy) = (dx)`

`(dy)/(dx) = 1/(2y+2)`

The equation of the line tangent to the graph at (0,3) is in the form,

`y = mx+c`

`m = (dy)/(dx) = 1/(2 xx 3+2) = 1/8`

`c = 3` (since when `x= 0, y =3` )

**Therefore the equation of the line is,**

`y = 1/8x+3`

**I am assuming that the second equation you have given is**

`y = 1/4x- 3`

Let `P = (x_1,y_1)` then,

The gradient `= 1/4`

Therefore,

`1/4 = 1/(2y_1+2)`

`1/2 = 1/(y_1+1)`

`2 = y_1 +1`

`y_1 = 1`

When `y_1 = 1, x_1` from the line is,` `

`1 = 1/4 x_1-3`

`x_1 = 16`

`x_1` from the graph is,

`1 -x_1+2-3=0`

`x_1 = 0.`

**Two `x_1` values are are not equal. Therefore, the line `y =1/4x -3` is not a tangent to the curve.**