The equation of the parabola with vertex at (-1,1) and focus (2,1) is :

Question

The equation of the parabola with vertex at (-1,1) and focus (2,1) is : (A) y2 - 2y - 12x +13 = 0 (B) y2 - 2y + 12x +11 = 0 (C) x2 + 2x - 12y +13 = 0 (D) y2 - 2y -12x -11 = 0

Tardigrade Admin · Accepted Answer

By the condition of parabola PM2 = PS2 ⇒ (x + 4)2 = (x-2)2 +(y - 1)2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/93ef056c1b9d49cde8b69dd20736676d-.png" /> ⇒ x2 + 8x - 16 =x2 - 4x + 4 + y2 - 2y + 1 ⇒ y2 - 2y - 12x - 11 = 0

Q. The equation of the parabola with vertex at $(- 1, 1)$ and focus $(2, 1)$ is :

$y^{2} - 2 y - 12 x + 13 = 0$

$y^{2} - 2 y + 12 x + 11 = 0$

$x^{2} + 2 x - 12 y + 13 = 0$

$y^{2} - 2 y - 12 x - 11 = 0$

By the condition of parabola
$P M^{2} = P S^{2}$
$\Rightarrow (x + 4)^{2} = (x - 2)^{2} + (y - 1)^{2}$

$\Rightarrow x^{2} + 8 x - 16 = x^{2} - 4 x + 4 + y^{2} - 2 y + 1$
$\Rightarrow y^{2} - 2 y - 12 x - 11 = 0$

Q. The equation of the parabola with vertex at (−1,1) and focus (2,1) is :

y2−2y−12x+13=0

y2−2y+12x+11=0

x2+2x−12y+13=0

y2−2y−12x−11=0