Question

The distance between the vertex of the parabola $ y={{x}^{2}}-4x+3 $ and the centre of the circle $ {{x}^{2}}=9-{{(y-3)}^{2}} $ is

• A. $ 2\sqrt{3} $
• B. $ 3\sqrt{2} $
• C. $ 2\sqrt{2} $
• D. $ \sqrt{2} $
• E. $ 2\sqrt{5} $

Answer 1

Answer: (E)

Given equation of parabola, $ y={{x}^{2}}-4x+3 $
$ \Rightarrow $ $ (y+1)={{x}^{2}}-4x+4 $
$ \Rightarrow $ $ {{(x-2)}^{2}}=(y+1) $ Comparing with, $ {{X}^{2}}=4aY $ here $ X=x-2 $ and $ y=y+1,a=\frac{1}{4} $ Vertex is $ (2,-1)=V $ (say) equation of circle, $ {{x}^{2}}=9-{{(y-3)}^{2}} $ or $ {{(x-0)}^{2}}+{{(y-3)}^{2}}=9 $ Centre $ C=(0,\text{ }3) $ Now, $ VC=\sqrt{{{(2-0)}^{2}}+{{(-1-3)}^{2}}} $
$=\sqrt{4+16}=\sqrt{20} $
$=2\sqrt{5} $