Question

What is the focus of the following graph?

Answer 1

@Michele_Laino

Answer 2

we have to write the equation of your parabola, first

Answer 3

the requested equation is:
y=-ax^2, where a>0
what is a?

Answer 4

\[\Large y = - a{x^2}\]

Answer 5

we know that your parabola passes at points:
(-2,-1) and (2,-1)

Answer 6

it would be 0 correct

Answer 7

substituting the coordinates of the first point into my formula above, we have:
\[\Large - 1 = - a{\left( { - 2} \right)^2}\]
so what is a?

Answer 8

a= 1

Answer 9

hint:
\[ - 1 = - a \times 4\]

Answer 10

is a=1/4 ?

Answer 11

yes i came up with that also

Answer 12

ok! so the equation of our parabola is:
\[y = - \frac{{{x^2}}}{4}\]

Answer 13

now, for a generic parabola:
\[y = M{x^2} + Px + Q\]
the coordinates of the focus, are:

\[\left\{ \begin{gathered}
{x_v} = - \frac{P}{{2M}} \hfill \\
{y_v} = \frac{{1 - {P^2} + 4MQ}}{{4M}} \hfill \\
\end{gathered} \right.\]

Answer 14

now you have to compare our parabola:
\[y = - \frac{{{x^2}}}{4}\]
with this one:
\[y = M{x^2} + Px + Q\]
what are the coefficients
M, P, and Q?

Answer 15

hmm

Answer 16

hint:
we can rewrite our parabola as below:
\[y = - \frac{{{x^2}}}{4} = \left( { - \frac{1}{4}} \right){x^2} + 0 \cdot x + 0\]

Answer 17

is Q=0?

Answer 18

yes. 0,0 is the line of symmetry for the parabola

Answer 19

we have:
\[M = - \frac{1}{4},\quad P = 0,\quad Q = 0\]
am I right?

Answer 20

yes

Answer 21

ok! Now substitute those values into these formulas:
\[\left\{ \begin{gathered}
{x_v} = - \frac{P}{{2M}} \hfill \\
{y_v} = \frac{{1 - {P^2} + 4MQ}}{{4M}} \hfill \\
\end{gathered} \right.\]
what do you get?

Answer 22

\[\Large \left\{ \begin{gathered}
{x_F} = - \frac{P}{{2M}} \hfill \\
{y_F} = \frac{{1 - {P^2} + 4MQ}}{{4M}} \hfill \\
\end{gathered} \right.\]

Answer 23

x=-0/2(-1/4)

Answer 24

x= - 0/-1/2

Answer 25

please simplify

Answer 26

x=...?

Answer 27

1/2

Answer 28

are you sure?
0/(-1/2)=0

Answer 29

\[{x_F} = - \frac{P}{{2M}} = - \frac{0}{{\left( { - 1/2} \right)}} = 0\]

Answer 30

x=0

Answer 31

ok! now, please compute y

Answer 32

hint:
\[{y_F} = \frac{{1 - {P^2} + 4MQ}}{{4M}} = \frac{{1 - {0^2} + 4 \times 0 \times 0}}{{4 \times \left( { - 1/2} \right)}} = ...?\]

Answer 33

y=1

Answer 34

\[\frac{{1 - {0^2} + 4 \times 0 \times 0}}{{4 \times \left( { - 1/2} \right)}} = \frac{{1 - 0}}{{4 \times \left( { - 1/2} \right)}} = ......?\]

Answer 35

y=-1 @Michele_Laino

Answer 36

are you sure?
\[\frac{{1 - {0^2} + 4 \times 0 \times 0}}{{4 \times \left( { - 1/2} \right)}} = \frac{{1 - 0}}{{4 \times \left( { - 1/2} \right)}} = \frac{1}{{ - 2}}\]

Answer 37

so the ansewr is D (0,-1)

Answer 38

yes im sure because M= - 1/4 not 1/2

Answer 39

Sorry! Sorry! you are right!:
\[\begin{gathered}
{y_F} = \frac{{1 - {P^2} + 4MQ}}{{4M}} = \frac{{1 - {0^2} + 4 \times 0 \times 0}}{{4 \times \left( { - 1/4} \right)}} = ...? \hfill \\
\hfill \\
\frac{{1 - {0^2} + 4 \times 0 \times 0}}{{4 \times \left( { - 1/4} \right)}} = \frac{{1 - 0}}{{4 \times \left( { - 1/4} \right)}} = \frac{1}{{ - 1}} \hfill \\
\end{gathered} \]

Answer 40

wow i understood thise. thanks

Answer 41

So the focus is the subsequent point:
\[F = \left( {0, - 1} \right)\]
Sorry again for my big error!

Answer 42

no problem ur a great person. @Michele_Laino

Answer 43

from the graph of your parabola, we can model your parabola with this equation:
\[y = a{\left( {x - 1} \right)^2} + 4\]

Answer 44

thanks for your compliment! :)

Answer 45

another way to derive the requested equation, is to note that the equation of the directrix is:
y=3
and the coordinate of the vertex are:
x=1, and y=4

Answer 46

now we have to refer to a general formula, like this:
\[y = a{x^2} + bx + c\]

Answer 47

the directrix of such parabola has the subsequent equation:
\[y = - \frac{{1 + {b^2} - 4ac}}{{4a}}\]

Answer 48

whereas the vertex has the subsequent coordinates:
\[V = \left( { - \frac{b}{{2a}},\;\frac{{1 - {b^2} + 4ac}}{{4a}}} \right)\]

Answer 49

so, using our data, we can write this system:
\[\left\{ \begin{gathered}
- \frac{{1 + {b^2} - 4ac}}{{4a}} = 3 \hfill \\
\hfill \\
- \frac{b}{{2a}} = 1 \hfill \\
\hfill \\
\frac{{1 - {b^2} + 4ac}}{{4a}} = 4 \hfill \\
\end{gathered} \right.\]

Answer 50

and we can try to solve it.

Answer 51

so, which method do you prefer?

Answer 52

the first one which uses this equation:
\[y = a{\left( {x - 1} \right)^2} + 4\]
and we have to explain how we got that equation, or
the second one in which we have to solve this algebraic system:
\[\left\{ \begin{gathered}
- \frac{{1 + {b^2} - 4ac}}{{4a}} = 3 \hfill \\
\hfill \\
- \frac{b}{{2a}} = 1 \hfill \\
\hfill \\
\frac{{1 - {b^2} + 4ac}}{{4a}} = 4 \hfill \\
\end{gathered} \right.\]

Answer 53

easiest

Answer 54

ok! The we will use the first one

Answer 55

Now we make this traslation:
\[\left\{ \begin{gathered}
x - 1 = X \hfill \\
y - 4 = Y \hfill \\
\end{gathered} \right.\]
where Y and X are the new coordinates

Answer 56

if we set X=0, we get x=...?

Answer 57

-1

Answer 58

x-1=0, from which x=...?

Answer 59

x=-1 @Michele_Laino

Answer 60

ok!

Answer 61

similarly, if I set Y=0, what is y?

Answer 62

-4

Answer 63

y-4=0, so y=...?

Answer 64

-4 @Michele_Laino

Answer 65

y=4

Answer 66

-4 correct?

Answer 67

\[\Large y - 4 = 0\]

Answer 68

no, since y-4=0, means y=4