Can someone please help me
1 answer:
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(a) ET = 18 and WE = 12
(b) FA = 16/3
HF = 8/3
ET = 8
WE = 4
<u>Explanation:</u>
WT = 30 ( WE + ET = 30)
WE = 30 - ET
2FA = 3HF
FA = 3/2HF
WE =?, ET = ?
Line WH, EF and TA are parallel to each other.
(a)
ΔWTA and ΔWHA are congruent to each other
Therefore, by congruency
So,
30 - ET/ ET = HF / 3/2HF
30 - ET / ET = 2HF / 3HF
30 - ET / ET = 2/3
On solving the above equation we get, ET = 18
WE = 30 - ET
WE = 30 - 18 = 12
Therefore, ET = 18 and WE = 12
(b)
HA = 8 ( HF + FA = 8)
HF = 8 - FA
WT = 12 ( WE + ET = 12 )
WE = 12 - ET
GA = 2WG
HF, WA, WE, ET = ?
ΔWHA and ΔGFA are congruent
So,
8 - FA / FA = WG / 2 WG
On solving the above equation we get,
FA = 16/3
HF + FA = 8
HF + 16/3 = 8
HF = 8/3
ΔWTA and ΔWEG are congruent
So,
12 - ET / ET = WG / 2WG
On solving the above equation, ET = 8
WE = 12 - ET
WE = 12 - 8
WE = 4
Therefore, FA = 16/3
HF = 8/3
ET = 8
WE = 4
If the focus is at (6, 2) and the directrix is a horizontal line y = 1, then that tells us that is an x^2 parabola. Since the parabola hugs the focus, it will open upwards since the focus is above the directrix. The rule here is that the vertex is the same distance from the focus as it is from the directrix. If the focus is at a y-value of 2 and the directrix is at y = 1, then the vertex is right in between them as far as the y coordinate goes, which is 1.5. It will have the same x coordinate at the focus. The vertex is in the form (h, k), so our h is 6, and our k is 1.5. The vertex is (6, 1.5). The standard form of a parabola of this type is 
, where p is the distance from the vertex to the focus. Our p is 1/2. Using the h and k from the vertex, and the p of 1/2, we now have this as our equation, not yet simplified: 
. That will simplify a bit to 
. Depending upon how you are to state your answer, how it needs to "look" in the end will vary. I am going to FOIL the left and distribute the right and then put everything on one side and set it equal to y. That would be this: 
. And there you go!