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May 14th, 2015, 07:43 AM   #1
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Parabola and Tangents

Hi, can anyone help me solve this problem? thanx a lot
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May 14th, 2015, 10:18 PM   #2
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There's a lot of algebra in solving part 1), but basically you know that the equation of line $BA$ is $y=-2x+7$. That line intersects the parabola at one point which is found by equating the two $y's$ or

$\displaystyle-2x+7=-x^2+ax+b\Longrightarrow x^2+(-2-a)x+(7-b)=0$

Realizing that the $x$-intersection point is positive in value (look at a graph of $y=-2x+7$), you get by the quadratic equation that

$\displaystyle{x}=\frac{2+a\color{red}{+}\sqrt{a^2 +4a+4b-24}}{2}$

Likewise, for line $BC$, the equation of the line is $y=6x+7$. That line also intersects the parabola at one point, whose $x$-value is found by solving

$\displaystyle6x+7=-x^2+ax+b\Longrightarrow x^2+(6-a)x+(7-b)=0$

This time, the $x$-intersection point has to be negative (look at a graph of $y=6x+7$), so the quadratic equation gives

$\displaystyle{x}=\frac{a-6\color{red}{-}\sqrt{a^2-12a+4b+8}}{2}$



These two $x$-intersection points can be used in the derivative of the parabola:

$\displaystyle\frac{dy}{dx}=-2x+a$

When the slope is $-2$ (i.e., line $BA$), then $\displaystyle{x}_{BA}=\frac{2+a\color{red}{+} \sqrt {a^2+4a+4b-24}}{2}$. Therefore,

$\displaystyle-2=(-2)\frac{2+a\color{red}{+}\sqrt{a^2+4a+4b-24}}{2}+a\Longrightarrow0=a^2+4a+4b-24$

When the slope is $6$ (i.e., line $BC$), then $\displaystyle{x}_{BC}=\frac{a-6\color{red}{-}\sqrt{a^2-12a+4b+8}}{2}$. Therefore,

$\displaystyle6=(-2)\frac{a-6\color{red}{-}\sqrt{a^2-12a+4b+8}}{2}+a\Longrightarrow0=a^2-12a+4b+8$


Now you have two variables $a$ and $b$ in two equations. I will let you do the algebra, but you should get

$a=2$ and $b=3$


Part 2) should be a little simpler.
Thanks from matisolla

Last edited by limiTS; May 14th, 2015 at 10:24 PM.
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November 28th, 2015, 02:18 AM   #3
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(1) The discriminants must be zero, so a² + 4a + 4b - 24 = a² - 12a + 4b + 8 = 0. Hence a = 2, and so b = 3.

(2) The parabolic part of the domain's boundary has "base length" 2 and height 1,
$\ \,\,\,\,\,$ so the required area = 2(4)/2 - (2/3)(2)(1) = 8/3.

Last edited by skipjack; November 28th, 2015 at 09:06 AM.
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November 28th, 2015, 03:04 AM   #4
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Find the equation of parabola, given vertex at (1,2) and focus at (3,4) ????

Last edited by skipjack; November 28th, 2015 at 09:03 AM.
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November 28th, 2015, 09:04 AM   #5
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What title would you like for your new subject?

Use the formula given here.
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