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: Find the equation of the parabola if the curve is open rightward, vertex is (2, 1) and passing through point (6, 5).

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 Solution:: The General equation of parabola whose vertex at (h ,k ) and open rightward is.

                 \ Rightarrow (y-k)^2=4a(x-h) 

We have,vertex  \=(h,k)=(2,1)

Hence 

The equation of the curve 

\Rightarrow (y-1)^2=4a(x-2)..........(1)

Parabola (1) passing through (6,5)

Now ,

\Rightarrow (5-1)^2=4a(6-2)Rightarrow 16=4a	imes 4\ \Rightarrow a=1

The equationof parabola will be 

\Rightarrow (y-1)^2=4(x-2)

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Deependra Verma

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