If the tangent to the conic y - 6 = x^2 at (2 | vedclass

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(D) Given the conic $y - 6 = x^2$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At the point $(2, 10)$,the slope of the tangent is

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$\left( - \frac{8}{17}, \frac{2}{17} \right)$

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(D) Given the conic $y - 6 = x^2$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At the point $(2, 10)$,the slope of the tangent is $m = 2(2) = 4$.The equation of the tangent at $(2, 10)$ is $y - 10 = 4(x - 2)$,which simplifies to $4x - y + 2 = 0$.Since the tangent touches the circle $x^2 + y^2 + 8x - 2y = k$ at $(\alpha, \beta)$,the point $(\alpha, \beta)$ lies on the tangent line,so $4\alpha - \beta + 2 = 0$,or $\beta = 4\alpha + 2$.The slope of the tangent to the circle at $(\alpha, \beta)$ is given by differentiating $x^2 + y^2 + 8x - 2y = k$,which gives $2x + 2y \frac{dy}{dx} + 8 - 2 \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{-(x + 4)}{y - 1}$.At $(\alpha, \beta)$,the slope is $4$,so $\frac{-(\alpha + 4)}{\beta - 1} = 4$,which implies $-\alpha - 4 = 4\beta - 4$,or $\alpha = -4\beta$.Substituting $\alpha = -4\beta$ into $\beta = 4\alpha + 2$,we get $\beta = 4(-4\beta) + 2$,so $\beta = -16\beta + 2$,which gives $17\beta = 2$,so $\beta = \frac{2}{17}$.Then $\alpha = -4 \left( \frac{2}{17} \right) = -\frac{8}{17}$.Thus,the point $(\alpha, \beta)$ is $\left( -\frac{8}{17}, \frac{2}{17} \right)$.

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

If the tangent to the conic $y - 6 = x^2$ at $(2, 10)$ touches the circle $x^2 + y^2 + 8x - 2y = k$ (for some fixed $k$) at a point $(\alpha, \beta)$,then $(\alpha, \beta)$ is

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