If (2,3) is the focus and x-y+3=0 is the dire | vedclass

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(B) Let the foot of the perpendicular from the focus $(2,3)$ to the directrix $x-y+3=0$ be $(h, k)$.Since the line passing through the focus and perpe

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$x-y+2=0$

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(B) Let the foot of the perpendicular from the focus $(2,3)$ to the directrix $x-y+3=0$ be $(h, k)$.Since the line passing through the focus and perpendicular to the directrix has the equation $\frac{x-2}{1} = \frac{y-3}{-1} = \lambda$,we have $x = 2+\lambda$ and $y = 3-\lambda$.Substituting into the directrix equation: $(2+\lambda) - (3-\lambda) + 3 = 0$ $\Rightarrow 2\lambda + 2 = 0$ $\Rightarrow \lambda = -1$.Thus,the foot of the perpendicular is $(2-1, 3+1) = (1, 4)$.The vertex is the midpoint of the focus $(2,3)$ and the foot of the perpendicular $(1,4)$,which is $(\frac{2+1}{2}, \frac{3+4}{2}) = (\frac{3}{2}, \frac{7}{2})$.The tangent at the vertex is parallel to the directrix $x-y+3=0$,so its equation is $x-y+c=0$.Substituting the vertex $(\frac{3}{2}, \frac{7}{2})$: $\frac{3}{2} - \frac{7}{2} + c = 0$ $\Rightarrow -2 + c = 0$ $\Rightarrow c = 2$.Therefore,the equation of the tangent at the vertex is $x-y+2=0$.

If $(2,3)$ is the focus and $x-y+3=0$ is the directrix of a parabola,then the equation of the tangent drawn at the vertex of the parabola is

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