If (2,3) is the vertex and (3,2) is the focus | vedclass

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(C) Given vertex $O = (2,3)$ and focus $S = (3,2)$.Let the directrix intersect the axis at point $A = (x_1, y_1)$. Since the vertex $O$ is the midpoin

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$x^2+2xy+y^2-18x-2y+17=0$

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(C) Given vertex $O = (2,3)$ and focus $S = (3,2)$.Let the directrix intersect the axis at point $A = (x_1, y_1)$. Since the vertex $O$ is the midpoint of $AS$,we have:$\frac{x_1+3}{2} = 2 \Rightarrow x_1 = 1$$\frac{y_1+2}{2} = 3 \Rightarrow y_1 = 4$So,$A = (1,4)$.The slope of the axis $AS$ is $m_1 = \frac{2-3}{3-2} = -1$.The directrix is perpendicular to the axis,so its slope $m_2 = -\frac{1}{m_1} = 1$.The equation of the directrix passing through $(1,4)$ with slope $1$ is:$y-4 = 1(x-1) \Rightarrow y-x-3 = 0$.By the definition of a parabola,for any point $P(x,y)$ on it,the distance to the focus equals the distance to the directrix:$PS^2 = PM^2$$(x-3)^2 + (y-2)^2 = \left(\frac{|x-y+3|}{\sqrt{1^2+(-1)^2}}\right)^2$$(x^2-6x+9) + (y^2-4y+4) = \frac{(x-y+3)^2}{2}$$2(x^2+y^2-6x-4y+13) = x^2+y^2+9-2xy+6x-6y$$x^2+2xy+y^2-18x-2y+17 = 0$.

If $(2,3)$ is the vertex and $(3,2)$ is the focus of a parabola,then its equation is

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