The focal chord of the parabola (y - 2)^2 = 1 | vedclass

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(B) The given parabola is $(y - 2)^2 = 16(x - 1)$. Comparing with $(y - k)^2 = 4a(x - h)$,we get $h = 1, k = 2$ and $4a = 16$,so $a = 4$.The focus of

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$1$

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(B) The given parabola is $(y - 2)^2 = 16(x - 1)$. Comparing with $(y - k)^2 = 4a(x - h)$,we get $h = 1, k = 2$ and $4a = 16$,so $a = 4$.The focus of the parabola is $(h + a, k) = (1 + 4, 2) = (5, 2)$.Any line passing through the focus $(5, 2)$ with slope $m$ is given by $(y - 2) = m(x - 5)$,which simplifies to $mx - y + (2 - 5m) = 0$.The given circle is $x^2 + y^2 - 14x - 4y + 51 = 0$. Its center is $(7, 2)$ and radius $r = \sqrt{7^2 + 2^2 - 51} = \sqrt{49 + 4 - 51} = \sqrt{2}$.Since the line is a tangent to the circle,the perpendicular distance from the center $(7, 2)$ to the line $mx - y + (2 - 5m) = 0$ must be equal to the radius $\sqrt{2}$.$\frac{|m(7) - 2 + 2 - 5m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$$4m^2 = 2m^2 + 2$$2m^2 = 2$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.Thus,the slope can be $1$ or $-1$.

The focal chord of the parabola $(y - 2)^2 = 16(x - 1)$ is a tangent to the circle $x^2 + y^2 - 14x - 4y + 51 = 0$. Then the slope of the focal chord can be:

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