A(-2,9) and B(1,6) are two points on the curv | vedclass

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(A) Given the curve $y=x^2+5$. Finding the slope of the tangent at $A(-2,9)$: $y' = 2x$. At $x = -2$,the slope $m = 2(-2) = -4$. Since the tangent at

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$(-5,30)$

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(A) Given the curve $y=x^2+5$. Finding the slope of the tangent at $A(-2,9)$: $y' = 2x$. At $x = -2$,the slope $m = 2(-2) = -4$. Since the tangent at $A$ is parallel to the chord $BC$,the slope of chord $BC$ must also be $-4$. Let $C$ be $(x', y')$ where $y' = x'^2 + 5$. The slope of chord $BC$ is $\frac{y' - 6}{x' - 1} = -4$. $y' - 6 = -4(x' - 1)$ $\Rightarrow y' - 6 = -4x' + 4$ $\Rightarrow y' = -4x' + 10$. Substituting $y' = x'^2 + 5$: $x'^2 + 5 = -4x' + 10 \Rightarrow x'^2 + 4x' - 5 = 0$. $(x' + 5)(x' - 1) = 0$. So,$x' = -5$ or $x' = 1$. If $x' = 1$,$C$ is $(1, 6)$,which is point $B$. If $x' = -5$,$y' = (-5)^2 + 5 = 30$. Thus,the coordinates of $C$ are $(-5, 30)$.

$A(-2,9)$ and $B(1,6)$ are two points on the curve $y=x^2+5$. The coordinates of the point $C$ on the curve such that the tangent drawn at $A$ is parallel to the chord $BC$,are

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