SOLUTION: Find the vertex and focus and end points of the latus rectum of the parabola 5x^2-5x+7y-4=0

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Question 1018044: Find the vertex and focus and end points of the latus rectum of the parabola 5x^2-5x+7y-4=0
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
7y=-5x%5E2%2B5x%2B4
7y=-5%28x%5E2-x%29%2B4
7y=-5%28x%5E2-x%2B1%2F4-1%2F4%29%2B4, completing the square using 1%2F4;
which, through a few more steps, will give
as the standard form, highlight%28y=-%285%2F7%29%28x-1%2F2%29%5E2%2B11%2F28%29
Telling you that the vertex, being a maximum, is at VERTEX: (1/2, 11/28).

The equation in an equivalent form is also highlight%28%287%2F5%29y=-%28x-1%2F2%29%5E2%2B11%2F20%29.
Comparing to 4py=-%28x-1%2F2%29%5E2%2B11%2F20, the value of p is the distance between the FOCUS and the vertex.
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4p=7%2F5
highlight_green%28p=7%2F20%29.

The focus, being below the vertex for this example, will have x=1/2 but y=11%2F28-7%2F20=%285%2F5%29%2811%2F28%29-%287%2F7%29%287%2F20%29
y=%2855-49%29%2F%28140%29
y=6%2F140
y=3%2F70
FOCUS: ( 1/2, 3/70 ).