SOLUTION: y=-2x^2-16x-35 find the vertex find the equation of the axis of symetry find the coordinates of the focus find the equation of the directrix find the length of the latus rect

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: y=-2x^2-16x-35 find the vertex find the equation of the axis of symetry find the coordinates of the focus find the equation of the directrix find the length of the latus rect      Log On


   



Question 63780: y=-2x^2-16x-35
find the vertex
find the equation of the axis of symetry
find the coordinates of the focus
find the equation of the directrix
find the length of the latus rectum

Answer by venugopalramana(3286) About Me  (Show Source):
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SEE THE FOLLOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
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(2) What is the (a)directrix and the (b)focus of the parabola y = x^2 - 5x + 4?
Y = [X^2-2*X*(5/2)+(5/2)^2]-(5/2)^2+4
(X-2.5)^2 = (Y+2.25)
COMPARING WITH STD. EQN. OF PARABOLA
(X-H)^2 = 4A(Y-K)
4A=1......A=1/4=0.25
VERTEX = (H,K)......= (2.5,-2.25)
AXIS = X-H=0.......X-2.5=0
FOCUS = [H,K+A].....[2.5,-2.25+0.25]=(2.5,-2)
DIRECTRIX = Y-K+A=0......Y+2.25+0.25=0.......Y+2.5=0