Find the vertex focus Axis directrix and the lattis rectum of the following parabola y^2=8x+8y.

Open in App

Solution

y2 =8x-8y y^2+8y=8x complete the square: y^2+8y+16=8x+16 (y+4)^2=8(x+2) This is an equation of a parabola that opens rightward. Its basic equation: (y-k)^2=4p(x-h) vertex: (-2,-4) axis of symmetry: y=-4 4p=8 p=2 focus: (0,-4) (p-distance to the right of the vertex on the axis of symmetry) directrix: x=-4 (p-distance to the left of the vertex on the axis of symmetry) latus rectum or focal width=4p=8

Suggest Corrections

Similar questions

y^{2} = 8 x + 8 y

y^{2} + 8 x - 6 y + 1 = 0

9 y^{2} - 16 x - 12 y - 57 = 0

y^{2} + 4 x + 4 y - 3 = 0

x^{2} - 2 x + 8 y + 17 = 0

Join BYJU'S Learning Program