If (h, k) is the point to which the origin is | vedclass

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(D) The given equation is $y^2-4x+6y+17=0$. Completing the square for the $y$ terms: $(y^2+6y+9)-9-4x+17=0$ $(y+3)^2-4x+8=0$ $(y+3)^2=4x-8$ $(y+3)^2=4

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(D) The given equation is $y^2-4x+6y+17=0$. Completing the square for the $y$ terms: $(y^2+6y+9)-9-4x+17=0$ $(y+3)^2-4x+8=0$ $(y+3)^2=4x-8$ $(y+3)^2=4(x-2)$. Comparing this with the standard form $Y^2=4aX$,where $Y=y+3$ and $X=x-2$,we see that the origin must be shifted to $(h, k) = (2, -3)$. Thus,$h=2$ and $k=-3$. Calculating $h^2+k^2$: $h^2+k^2 = (2)^2+(-3)^2 = 4+9 = 13$. Therefore,the correct option is $D$.

If $(h, k)$ is the point to which the origin is shifted in order to transform the equation $y^2-4x+6y+17=0$ into the form $Y^2=4aX$,then $h^2+k^2=$

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