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11 Jul 2004, 18:58
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Please explain.
(I do not know the method to upload the way BHAI does.) Can someone edit this and show the graph, instead of the download option). Or perhaps, let me know how to show the graph on the screen.
Thanks
(I do not know the method to upload the way BHAI does.) Can someone edit this and show the graph, instead of the download option). Or perhaps, let me know how to show the graph on the screen.
Thanks
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11 Jul 2004, 20:34
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The diagram is not so clear and not to scale.
Anyways here's my approach:
Eq of a parabola with vertex at (x0,y0) is
(y-y0)^2=4a(x-x0)^2 (x,y) any pint on the parabola. 4a is called the latus rectum (it is just a constant)
Based on the diagram looks like the parabola passes thru (0,3)
vertex of the parabola here is (2,0)
Using this we get
9=4a*4
or a=9/16
hence eq is (y)^2 = 9/4 * (x-2)^2
when, x=3 y=3/2=1.5
This is the answer I get, though its not in the answer choice.
Did I commit any mistake?
Anyways here's my approach:
Eq of a parabola with vertex at (x0,y0) is
(y-y0)^2=4a(x-x0)^2 (x,y) any pint on the parabola. 4a is called the latus rectum (it is just a constant)
Based on the diagram looks like the parabola passes thru (0,3)
vertex of the parabola here is (2,0)
Using this we get
9=4a*4
or a=9/16
hence eq is (y)^2 = 9/4 * (x-2)^2
when, x=3 y=3/2=1.5
This is the answer I get, though its not in the answer choice.
Did I commit any mistake?
