SOLUTION: Find the latus rectum, vertex and directrix of the parabola y^2-8x-2y+17=0

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Question 1129284: Find the latus rectum, vertex and directrix of the parabola y^2-8x-2y+17=0
Found 3 solutions by MathLover1, josgarithmetic, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Find the latus rectum, vertex and directrix of the parabola
y%5E2-8x-2y%2B17=0+...first write it in vertex form
%28y%5E2-2y%2Bb%5E2%29-b%5E2-8x%2B17=0+
%28y%5E2-2y%2B1%5E2%29-1%5E2-8x%2B17=0+
%28y-1%29%5E2%29-8%28x-2%29=0+
%28y-1%29%5E2%29=8%28x-2%29=0+=> compare to form %28y-k%29%5E2=4p%28x-h%29
=>h=2 and k=1 => vertex is at (2, 1)
4p=8
p=2
focus: (h%2Bp,k)= (4,+1)
Equation of the directrix is x=-p
x-2=-2
x=2-2
x+=+0

The line segment that passes through the focus (4,+1) and is parallel to the directrix x=0 is called the latus rectum.
The endpoints of the latus rectum lie on the curve, and they are:
(p,2p) and (p,-2p)
since p=2, we have
(2,4) and (2,-4)
distance between them is the length of the latus rectum:
d=sqrt%28%282-2%29%5E2%2B%284-%28-4%29%29%5E2%29
d=sqrt%280%2B%284%2B4%29%5E2%29
d=sqrt%288%5E2%29
d=8
so, length of the latus rectum = +8


Answer by josgarithmetic(39617) About Me  (Show Source):
Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find the latus rectum, vertex and directrix of the parabola y^2-8x-2y+17=0
This is a parabola with a HORIZONTAL AXIS of SYMMETRY, and so, its equation is: matrix%281%2C3%2C+%28y+-+k%29%5E2%2C+%22=%22%2C+4p%28x+-+h%29%29
Given equation: matrix%281%2C3%2C+y%5E2+-+8x+-+2y+%2B+17%2C+%22=%22%2C+0%29
After COMPLETING the SQUARE on y, we get:
Comparing that to the equation of a parabola with a HORIZONTAL AXIS of SYMMETRY, or matrix%281%2C3%2C+%28y+-+k%29%5E2%2C+%22=%22%2C+4p%28x+-+h%29%29, we see that:
h = 2; k = 1
4p = 8, or matrix%281%2C5%2C+p%2C+%22=%22%2C+8%2F4%2C+%22=%22%2C+2%29