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 Calculus Calculus Math Forum

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 Attached Images 11074540_10153338752554700_56516874_n.jpg (24.1 KB, 22 views) May 14th, 2015, 10:18 PM #2 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 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. November 28th, 2015, 02:18 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 (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. November 28th, 2015, 03:04 AM #4 Newbie   Joined: Nov 2015 From: hyderabad Posts: 1 Thanks: 0 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. November 28th, 2015, 09:04 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 What title would you like for your new subject? Use the formula given here. Thanks from Rishabh11199 Tags parabola, tangents Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hannah9896 Calculus 1 January 11th, 2015 08:26 AM Opposite Geometry 2 November 20th, 2014 04:08 AM Daltohn Algebra 8 October 23rd, 2013 02:37 AM Aspiring_Physicist Algebra 1 March 17th, 2013 04:25 AM suuup Calculus 2 October 31st, 2012 06:46 PM

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