1.a. state the parabola y^2 - 8x - 4y + 44 = 0 in conical form b. Find the I. Focus II. Directrix III. The coordinates of the ends of the Latus rectum?

1 Answer
Shwetank Mauria
Nov 5, 2017

(a) Canonical form of equation of parabola is #(x-5)=1/8(y-2)^2#
(b) I. Focus is #(7,2)#. II. Directrix is #x=3# and III. coordinates of end points of latus rectum are #(7,6)# and #(7,-2)#.

Explanation:

Canonical form of equation of parabola is #(y-k)=4a(x-h)^2# for vertical parabola and #(x-h)=4a(y-k)^2# for horizontal parabola. The vertex (in both cases) is #(h,k)#.

In the equation #(y-k)=a(x-h)^2#, focus is #(h,k+1/(4a))# and directrix is #y=k-1/(4a)# and in case equation is #(x-h)=a(y-k)^2#, focus is #(h+1/(4a),k)#, directrix is #x=h-1/(4a)#.

Points on the latus rectum are two times away from the focus than the distance of focus from vertex, but at right angles to the axis of symmetry.

The given equation is #y^2-8x-4y+44=0# and as squared term is #y# it is horizontal parabola.

#y^2-8x-4y+44=0# can be written as

#8x=y^2-4y+44=(y^2-4y+4)+40#

or #8x-40=(y-2)^2#

or #(x-5)=1/8(y-2)^2#

Hence, we have #h=5#, #k=2#, #a=1/8# and vertex is #(5,2)#. Therefore

I. Focus is #(5+1/(4xx1/8),2)# i.e. #(7,2)#. Note that its distance from focus is #2# units.

II. Directrix is #x=5-1/(4xx1/8)=3#

III. Coordinates of end points of latus rectum are #(7,6)# and #(7,-2)#.

graph{(y^2-8x-4y+44)(x-3)((x-7)^2+(y-2)^2-0.03)((x-7)^2+(y-6)^2-0.03)((x-7)^2+(y+2)^2-0.03)(y-2)(x-7)=0 [-1.29, 18.71, -3.08, 6.92]}

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