. Put the equation into the standard form of a parabola and find
the vertex, focus, and directrix (needs to be an equation of a
line). The standard form of a parabola can be found in Section 7.3
of the textbook. Answers must be exact values and not
approximations. 4y2 + 2 = -x + 8y
A. Standard form
B.Vertex
C.Focus
D.Directrix
identify the equation . If it is a parabola, give its vertex,
focus, and directrix; if an ellipse, give its centre, vertices, and
foci ; if a hyperbola, give its centre, vertices, foci, and
asymptotes.
(x + 1)^2/4 - y^2/9−=1
What is the standard form of the equation of the parabola with
the focus (−1,3) and the directrix x=9?
Select the correct
answer below:
a) (x+1)2=−12(y−6)
b) (y−3)2=−20(x−4)
c) (x−9)2=8(y−1)
d)(y−3)2=20(x−4)
please show work and what is the correct answer? thanks
ASAP
(Algebra: vertex form equations) The equation of a parabola can
be expressed in either standard form (y = ax^2 + bx + c) or vertex
form (y = a(x-h)^2 + k). Write a program that prompts the user to
enter a, b, and c as integers in standard form and displays h and k
in the vertex form. Hint: Use the Rational class in LiveExample
13.13 for computing h and k. Use the template at
https://liveexample.pearsoncmg.com/test/Exercise13_21Test.txt for
your code....
The equation of parabola is x2 = 4ay . Find the length of the
latus rectum of the parabola and length of the parabolic arc
intercepted by the latus rectum. (a) Length of latusrectum : 4a;
Length of the intercepted arc : 4.29a (b)Length of latusrectum:
—4a; Length of the intercepted arc : 4.39a (c) Length of
latusrectum: —4a; Length of the intercepted arc : 4.49a (d) Length
of latusrectum: 4a; Length of the intercepted arc : 4.59a