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

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 ,

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

The equationof parabola will be

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

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