A parabola is drawn with its focus at (3,4) and vertex at the focus of the parabola y2-12 x-4 y+4=0. The equation of the parabola is

Question

A parabola is drawn with its focus at (3,4) and vertex at the focus of the parabola y2-12 x-4 y+4=0. The equation of the parabola is (A) y2-8 x-6 y+25=0  (B) y2-6 x+8 y-25=0  (C)  x2-6 x-8 y+25=0  (D)  x2+6 x-8 y-25=0

Tardigrade Admin · Accepted Answer

Given equation can be rewritten as (y-2)2=12 x Here, vertex and foci are (0,2) and (3,2). ∴ Vertex of the required parabola is (3,2) and focus is (3,4). The axis of symmetry is x=3 and latusrectum =4 ⋅ 2=8 Hence, required equation is (x-3)2=8(y-2) or x2-6 x-8 y+25=0

Q. A parabola is drawn with its focus at $(3, 4)$ and vertex at the focus of the parabola $y^{2} - 12 x - 4 y + 4 = 0$ . The equation of the parabola is

$y^{2} - 8 x - 6 y + 25 = 0$

$y^{2} - 6 x + 8 y - 25 = 0$

$x^{2} - 6 x - 8 y + 25 = 0$

$x^{2} + 6 x - 8 y - 25 = 0$

Q. A parabola is drawn with its focus at (3,4) and vertex at the focus of the parabola y2−12x−4y+4=0. The equation of the parabola is

y2−8x−6y+25=0

y2−6x+8y−25=0

x2−6x−8y+25=0

x2+6x−8y−25=0

Given equation can be rewritten as (y−2)2=12x Here, vertex and foci are (0,2) and (3,2). ∴ Vertex of the required parabola is (3,2) and focus is (3,4). The axis of symmetry is x=3 and latusrectum =4⋅2=8 Hence, required equation is (x−3)2=8(y−2) or x2−6x−8y+25=0

Q. A parabola is drawn with its focus at $(3, 4)$ and vertex at the focus of the parabola $y^{2} - 12 x - 4 y + 4 = 0$ . The equation of the parabola is

$y^{2} - 8 x - 6 y + 25 = 0$

$y^{2} - 6 x + 8 y - 25 = 0$

$x^{2} - 6 x - 8 y + 25 = 0$

$x^{2} + 6 x - 8 y - 25 = 0$