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) x2-6 x-8 y+25=0 (B) y2-8 x-6 y+25=0 (C) x2-6 x+8 y-25=0 (D) x2+6 x-8 y-25=0

Tardigrade Admin · Accepted Answer

y2-12 x-4 y+4=0 y2-4 y=12 x-4 (y-2)2=12 x Y2=12 X focus: X=A, Y=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d046a8047d74e3187233540a2aeb6a27-.png" /> x=3, y=2 A(3,2) equ. of directrix is y=0, PS = PM √(x-3)2+(y-4)2=|y| by squaring, we will get (x-3)2+(y-4)2=y2 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:

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

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

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

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

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:

x2−6x−8y+25=0

y2−8x−6y+25=0

x2−6x+8y−25=0

x2+6x−8y−25=0

y2−12x−4y+4=0 y2−4y=12x−4 (y−2)2=12x Y2=12X focus : X=A,Y=0 x=3,y=2 A(3,2) equ. of directrix is y=0,PS=PM (x−3)2+(y−4)2​=∣y∣ by squaring, we will get (x−3)2+(y−4)2=y2 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:

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

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

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

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