find the vertex, focus, diretrix and length of latus-rectum of the parabola y2 - 4y - 2x - 8 - Maths - Conic Sections

Question

find the vertex, focus, diretrix and length of latus-rectum of
the parabola

y2 - 4y - 2x - 8 = 0
2y2 + 3y - 4x -3 = 0
9y2 - 16x - 12y - 57 = 0

Ajanta Trivedi · Accepted Answer

the given equation of parabola is:

$y^{2} - 4 y - 2 x - 8 = 0 y^{2} - 4 y + 4 = 2 x + 8 + 4 = 2 x + 12 (y - 2)^{2} = 2 (x + 6) ........... (1)$

shifting the origin to the point (-6,2) without rotating the axes and denoting the new coordinates with respect to these new axes by X and Y, we have

x=X-6 , y=Y+2 .....(2)

using these relations , equation (1) reduces to $Y^{2} = 2 X$ .....(3)

this is of the form $Y^{2} = 4 a X$ . on comparing we get 4a=2 ie. a=1/2

vertex: the coordinates of the vertex with respect to the new axes are X=0, Y=0

so, the coordinates of the vertex with respect to the old axes are (-6,2).

focus: the coordinates of the focus with respect to the new axes are (X=a, Y=0) i.e. (X=1/2, Y=0)

so, the coordinates of the focus with respect to the old axes are $(\frac{1}{2} - 6, 2) = (- \frac{11}{2}, 2)$

directrix: the equation of the directrix with respect to the new axes is X=-a i.e. $X = - \frac{1}{2}$

so, the equation of the directrix with respect to the old axes is $x = - \frac{1}{2} - 6 \Rightarrow x = - \frac{13}{2}$

latus rectum: the length of the latus-rectum of the parabola is equal to 4a=2

i hope you can solve (ii) and (iii) question in the similar way.

find the vertex, focus, diretrix and length of latus-rectum of the parabola y2 - 4y - 2x - 8 = 0 2y2 + 3y - 4x -3 = 0 9y2 - 16x - 12y - 57 = 0