What is the focus of the parabola x-4y^2+16y-19=0x−4y2+16y−19=0?

1 Answer
SagarStudy
Jun 13, 2016

The coordinates of focus of the given parabola are (49/16,2).(4916,2).

Explanation:

x-4y^2+16y-19=0x−4y2+16y−19=0
implies 4y^2-16y+16=x-3⇒4y2−16y+16=x−3
implies y^2-4y+4=x/4-3/4⇒y2−4y+4=x4−34
implies (y-2)^2=4*1/16(x-3)⇒(y−2)2=4⋅116(x−3)
This is a parabola along x-axis.
The general equation of a parabola along x-axis is (y-k)^2=4a(x-h)(y−k)2=4a(x−h),
where (h,k)(h,k) are coordinates of vertex and aa is the distance from vertex to the focus.

Comparing (y-2)^2=4*1/16(x-3)(y−2)2=4⋅116(x−3) to the general equation, we get

h=3, k=2h=3,k=2 and a=1/16a=116

implies⇒ Vertex=(3,2)Vertex=(3,2)

The coordinates of focus of a parabola along x-axis are given by (h+a,k)(h+a,k)

implies Focus=(3+1/16,2)=(49/16,2)⇒Focus=(3+116,2)=(4916,2)

Hence, the coordinates of focus of the given parabola are (49/16,2).(4916,2).

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