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 = x2 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.
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) 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
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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