Given the equation y^2 - x + 2y - 3=0....questions below?

Question

a. find dy/dx
b. Write an equation of the line tangent to the graph of the equation at the point (0, -3).
c. write an equation of the line normal to the graph of the equation at the point (0, -3)

Answer 1

a.
Implicit differentiation:

2y dy/dx - 1 + 2 dy/dx = 00

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

b.
y=mx+c = x/(2y+2) + c

at the point (0,-3):

-3 = c
=>
y = x/(-4) - 3

so

4y+x+12=0

c.

Same as (b), except the gradient is -1/m. Can you do that?

Answer 2

You can use completing the square to rewrite this as:
y^2 - x + 2y - 3 = 0
y^2 + 2y = x + 3
y^2 + 2y + 1 = x + 4
(y + 1)^2 = x + 4
y = -1±√(x+4)
So dy/dx = (±1/2)(x+4)^(-1/2)

You have one y^2 term in the original equation, so this is a parabola that opens either to the left or the right. This means there will be more than one y value for a given x value, so the ± sign makes sense here.

Another way is to take the derivative of (y + 1)^2 = x + 4 to get 2(y+1) dy/dx = 1, so dy/dx = 1 / 2(y+1). Therefore the slope of the tangent line at (0,-3) is 1 / 2(-3+1) = -1/4. With this slope and a point on the line, you should now be able to write an equation for the tangent line. The line normal to this is obviously the line perpendicular to this line at the same point. So the slope would be 4, and you can write an equation for the line.

Answer 3

In standard form (for parabolas) you have

(y+1)² = x+4,

which is a parabola opening right, vertex at (-4,-1), y intercepts at (0,1) and (0,-3).

Written as a pair of functions of x,

y = ±√(x+4) - 1, and so

dy/dx = ±1/[2√(x+4)], and at x=0,

dy/dx = -1/[2√4] = -1/4 (neg since bottom half of parabola)

So tangent line is y = (-1/4)x - 3, and normal is y = 4x - 3.

Answer 4

2yy' -1 +2y' = 0
y'(2y+2)= 1
y' = dy/dx = 1/(2y+2)
Slope at (0,-3) = 1/(-6+2) = -1/4
y=-x/4 +b
-3 = b, so y = -x/4 - 3 <-- Equation of tangent
y= 4x +b
y= 4x-3 <-- Equation of normal.