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Find the equation of parabola if : Vertex $(2 , 3 )$ open upward ;and passing through the point $(6 , 4 )$.

This is the nineth part of the multi-part question Q1

Toolbox:
• Standard parabolas :
• (i)$y^2=4ax$
• In this case the axis of symmetry is $x$-axis.
• The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
• (ii)$y^2=-4ax$
• In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
• (iii)$x^2=4ay$
• In this case the axis of symmetry is $y$-axis.
• The focus is $F(0,a)$ and the equation of directrix is $y=-a$
• (iv)$x^2=-4ay$
• In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
• Shifting of origin from (0,0) to $(h,k)$ then if the new axes form $X-Y$ coordinate system then
• $x=X+a$
• $y=Y+k$
• The new coordinates of a point $P(x,y)$ become $X=x-a$,$Y=y-k$.
Step 1:
The parabola opens upward,the equation of the form $X^2=4aY$
Where $X=(x-2),Y=y-3$
$(x-2)^2=4a(y-3)$
Step 2:
It passes through $P(6,4)$
Therefore $x,y\Rightarrow 6,4$
$(6-2)^2=4a(4-3)$
$4^2=4a(1)$
$16=4a$
$a=4$
The required equation is $(x-2)^2=4\times 4(y-3)$
$\Rightarrow (x-2)^2=16(y-3)$
edited Jun 16, 2013