1v=9. Which equation represents the equation of theparabola with focus (-3,3) and directrix y = 7?A),= (x+3)2 – 5B) y = 5(? – 3)2 +5C) y=-( + 3)2 +5D) y=-2 (2-3)2 +5

1v9 Which equation represents the equation of theparabola with focus 33 and directrix y 7A x32 5B y 5 32 5C y 32 5D y2 232 5 class=

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SOLUTION

From the focus (-3, 3) and the directrix y = 7 given, note that the vertex is usually between the focus and the directrix.

So, the vertex will have the same x-coordinate as the focus, which is -3, and the y-coordinate of the vertex becomes

[tex]\begin{gathered} \frac{3+7}{2} \\ that\text{ is 3 from the y-coordinate of the focus and 7 from the directrix} \\ y=7 \\ \frac{3+7}{2}=\frac{10}{2}=5 \end{gathered}[/tex]

Hence coordinate of the vertex is (-3, 5)

Now, equation of a parabola is given as

[tex]\begin{gathered} (x-h)^2=4p(y-k) \\ where\text{ \lparen h, k\rparen is the coordinate of the vertex and p is the focal length} \\ y=k-p,\text{ so we have } \\ 7=5-p \\ p=5-7=-2 \end{gathered}[/tex]

So putting in the values of h, k and p into the equation, we have

[tex]\begin{gathered} (x-h)^{2}=4p(y-k) \\ (x-(-3)^2=4(-2)(y-5) \\ (x+3)^2=-8(y-5) \\ -\frac{1}{8}(x+3)^2=y-5\text{ that is dividing through by -8} \\ making\text{ y the subject, we have } \\ y=-\frac{1}{8}(x+3)^2+5 \end{gathered}[/tex]

Hence the answer is

[tex]y=-\frac{1}{8}(x+3)^{2}+5[/tex]