Question
Question: A parabola is drawn with focus at \[\left( {3,4} \right)\] and vertex at the focus of another parabo...
A parabola is drawn with focus at (3,4) and vertex at the focus of another parabola y2−12x−4y+4=0. Then equation of the parabola is
A) x2−6x−8y+25=0
B) y2−8x−4y+28=0
C) x2−6x+8y−25=0
D) x2−4x−8y+28=0
Solution
First find the focus of the given parabola y2−12x−4y+4=0.
Also given that, the focus of the above parabola is the vertex of the asked parabola.
It is given that; A parabola is drawn with focus at (3,4) and vertex at the focus of another parabola y2−12x−4y+4=0.
Now, find the equation of the parabola by using formula (x−h)2=4a(y−k).
Complete step by step solution:
A parabola is drawn with focus at (3,4) and vertex at the focus of another parabola y2−12x−4y+4=0.
Now, the equation y2−12x−4y+4=0 can be written as
On comparing the above equation of parabola with (y−h)2=4a(x−k) , we get h=2,a=3 and k=0 .
So, the focus of the parabola will be (a,h)=(3,2) and vertex as (k,h)=(0,2)
Now, it is given that the parabola drawn has focus on the given parabola as its vertex.
So, the parabola which is to be drawn has focus (3,4) and vertex (3,2) .
So, its latus-rectum will be =4(2)=8 .
Thus, the equation of asked parabola can be given as
(x−3)2=8(y−2) ⇒x2−6x+9=8y−16 ⇒x2−6x−8y+25=0
So, option (A) is correct.
Note:
Parabola:
A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point which is known as the focus and from a fixed straight line which is known as the directrix.
The general equation of parabola if the directrix is parallel to the Y-axis in the standard equation of a parabola is given as y2=4ax.
While the general equation of parabola, if the directrix is parallel to the X-axis in the standard equation of a parabola, is given as x2=4by.