What is the focus of the parabola 4y^2 + 12x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the parabola is $4y^2 + 12x - 20y + 67 = 0$. Rearranging the terms,we get $4y^2 - 20y = -12x - 67$. Dividing by $4$,we have

Vedclass · Accepted Answer

$(-17/4, 5/2)$

Vedclass · Answer

(C) The given equation of the parabola is $4y^2 + 12x - 20y + 67 = 0$. Rearranging the terms,we get $4y^2 - 20y = -12x - 67$. Dividing by $4$,we have $y^2 - 5y = -3x - 67/4$. Completing the square on the left side: $(y - 5/2)^2 = -3x - 67/4 + 25/4$. $(y - 5/2)^2 = -3x - 42/4 = -3x - 21/2$. $(y - 5/2)^2 = -3(x + 7/2)$. Comparing this with the standard form $Y^2 = 4aX$,where $Y = y - 5/2$,$X = x + 7/2$,and $4a = -3$ (so $a = -3/4$). The focus of $Y^2 = 4aX$ is $(a, 0) = (-3/4, 0)$. Thus,$x + 7/2 = -3/4$ and $y - 5/2 = 0$. $x = -3/4 - 14/4 = -17/4$ and $y = 5/2$. The focus is $(-17/4, 5/2)$.

What is the focus of the parabola $4y^2 + 12x - 20y + 67 = 0$?

Similar Questions

The equation ${y^2} - 2x - 2y + 5 = 0$ represents:

If the focus of the parabola is $(3, -4)$ and the directrix is $y - 4 = 0,$ then the equation of the parabola is :-

Find the equation of the parabola with focus $(4, -3)$ and vertex $(4, -1)$.

If a normal drawn to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ meets the parabola again at $(at^2, 2at)$,then the value of $t$ is:

What are the parametric equations of the parabola $y^2 = 8x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module