1)Given V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix.

2) Find the tangent to the parabola y = x

^{2 }using the fact that the equation of the line tangent to a curve y = f(x) is given by:

y - y1 = f'(x1) (x - x1)

where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1 = (x1)

^{2 }. Show the graph of the given parabola and the tangent line with the

*point P1 (x1, y1) identified*.

^{ 2 }

**- 4 y**

^{ 2 }- 2 x + 8 y - 2 = 0

Solutions:

1) From the vertex coordinates we have: h = -3 and k = 1

With these values this indicates a parabola opening to the right with the general form:

(y - k)

applies, so:

(y - 1)

For which 4p = 12 and p = 3, so we need a directrix equation of form x = -p

With these values this indicates a parabola opening to the right with the general form:

(y - k)

^{2}= 4p (x - h)applies, so:

(y - 1)

^{2}= 12 (x + 3 )For which 4p = 12 and p = 3, so we need a directrix equation of form x = -p

The graph is shown below:

2) We have: f'(x1)
= dy/ dx = 2x

So: y - y1 = 2x1 (x - x1)

But only one point satisfies y1 = (x1)

^{2}

^{ }

That is x1 = 1, y1 = 1

Hence:

y - 1 = 2 (x -1) or: y = 2x - 1

And this tangent line is shown below for the point P1 (x1, y1) = (1,1)

3) Collect x and y terms separately:

(x

^{ 2 }**- 2 x) - 4 (y**^{ 2 }- 2 y) = 7
(x

^{ 2 }**- 2 x + 1) - 4(y**^{ 2 }- 2 y + 1) = 7 + 1 - 4 = 5
Or:

(x + 1 )

^{2 }/ 4 - (y - 1 )^{2 }= 1
Let: x’
= x + 1,
and: y’ = y - 1

Then: x’

^{2 }/ 4 - y’^{2 }/ 1 = 1
Graphing:

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