If L(p, q), q 3 is one end of the latus rec | vedclass

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(B) The given parabola is $(y-2)^2 = 3(x-1)$.Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=1, k=2$,and $4a=3$,so $a = \frac{3}{4}$.The coordinates

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$4x - 4y + 7 = 0$

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(B) The given parabola is $(y-2)^2 = 3(x-1)$.Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=1, k=2$,and $4a=3$,so $a = \frac{3}{4}$.The coordinates of the focus are $(h+a, k) = (1 + \frac{3}{4}, 2) = (\frac{7}{4}, 2)$.The ends of the latus rectum are $(h+a, k \pm 2a) = (\frac{7}{4}, 2 \pm \frac{3}{2})$.Thus,the ends are $(\frac{7}{4}, \frac{7}{2})$ and $(\frac{7}{4}, \frac{1}{2})$.Since $q > 3$,the point $L$ is $(\frac{7}{4}, \frac{7}{2})$.The equation of the tangent at $(x_1, y_1)$ to the parabola $(y-k)^2 = 4a(x-h)$ is $(y-k)(y_1-k) = 2a(x+x_1-2h)$.Substituting $x_1 = \frac{7}{4}, y_1 = \frac{7}{2}, h=1, k=2, a=\frac{3}{4}$:$(y-2)(\frac{7}{2}-2) = 2(\frac{3}{4})(x+\frac{7}{4}-2(1))$$(y-2)(\frac{3}{2}) = \frac{3}{2}(x-\frac{1}{4})$$y-2 = x-\frac{1}{4}$$x-y+\frac{7}{4} = 0$,which simplifies to $4x-4y+7=0$.

If $L(p, q), q > 3$ is one end of the latus rectum of the parabola $(y-2)^2 = 3(x-1)$,then the equation of the tangent at $L$ to this parabola is

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