The slope of the tangent to the circle (x-6)^ | vedclass

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(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$.The eq

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$\pm 1$

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(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$.The equation of a line passing through $(4, 0)$ with slope $m$ is $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.This line is a tangent to the circle $(x-6)^2 + y^2 = 2$. The center of the circle is $(6, 0)$ and its radius is $r = \sqrt{2}$.The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$:$\frac{|m(6) - 0 - 4m|}{\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$$2m^2 = 2$$m^2 = 1$$m = \pm 1$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

The slope of the tangent to the circle $(x-6)^2 + y^2 = 2$,which passes through the focus of the parabola $y^2 = 16x$,is:

Similar Questions

Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
$A. (7, 2)$ $1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$ $2. -8$
$C. (1, -1)$ $3. 4$
$D. (3, \sqrt{2})$ $4. 0$
$5. -2\sqrt{2}$

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