Grade 11Analytical Geometry
the parabola having its focus at (3,2) and directrix along the y-axis has its vertex at:
2 Answers
Shahid
Let the Equation of the Parabola be (y-k)2 = 4a(x-h)
It’s vertex is (h.k)
The Equation of Directrix for this parabola is (according to the equation) is x+a = h ------{1}
It’s Focus is (a+h,k) = (3,2)
Thus a+h = 3 ---------{2}
k = 2
Directrix along the y-axis … Therefore Equation of Directrix is x=0 -------------{3}
Now , {1} and {3} represents the same line...
Therefore , a = h
Substituting this in {2} , we get
2h = 3
h = 3/2
Now... Vertex = (h.k) = (3/2 ,2 )
RAVI KUMAR
(y-k)^2=4p(x-h)
Since,the vertex and focus are3 unit apart, we have p=3, and since the vertix is at(h,k)=1,2, we obtain
(y-2)^2=12(x-1)
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