Question

The axis of a parabola is along the line y=x and the distance of its vertex from origin is squrt2 and that from its focus is 2squrt2. If vertex and focus both lie in the first quadrant , then the equation of the parabola is…………………………?

Answer 1

 [tex]vertex = (1,1),\ \ \ \ focus = (1+2\sqrt{2}, 1+2\sqrt{2}),\ \ \ \ Axis: y - x = 0 \\ \\ Directrice\ =\ x + y - 2 + 4\sqrt{2} = 0 \\ \\ Distance\ from\ directrice\ =\ Distance\ from\ focus \\ \\ [/tex]

[tex] \frac{[x+y - 2 + 4 \sqrt{2}]^{2}}{1^2 + 1^2} = [ x - 1 - 2\sqrt{2} ]^{2} + [y - 1 - 2\sqrt{2}]^{2} \\ \\On\ simplification\ we\ get\ \\ \\ x^2 + y^2 - 2 x y - 16\sqrt{2} - 16 \sqrt{2} y + 32 \sqrt{2} = 0 \\ \\ [/tex]

Answer 2

Step-by-step explanation:

The parabola's vertex and focus lie on x-ais at points (a, 0) and (b, 0). Vertex and focus lie on the x-axis hence, the aixs of parabola is x-axis. Equation of parabola Vertex whose is a point(x1,y1) then is

(y−y1)2=4(x−x1)

So, y1=0 and x1=aand k = distance between focus and vetex = (b-a) so the equation is

(y−0)2=4(b−a).(x−a)

i.e., y2=4(b−a)(x−a)