May 14th, 2015, 07:43 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Parabola and Tangents
Hi, can anyone help me solve this problem? thanx a lot 
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 $\displaystyle2x+7=x^2+ax+b\Longrightarrow x^2+(2a)x+(7b)=0$ 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 $\displaystyle{x}=\frac{2+a\color{red}{+}\sqrt{a^2 +4a+4b24}}{2}$ 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 $\displaystyle6x+7=x^2+ax+b\Longrightarrow x^2+(6a)x+(7b)=0$ This time, the $x$intersection point has to be negative (look at a graph of $y=6x+7$), so the quadratic equation gives $\displaystyle{x}=\frac{a6\color{red}{}\sqrt{a^212a+4b+8}}{2}$ 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 $\displaystyle{x}_{BA}=\frac{2+a\color{red}{+} \sqrt {a^2+4a+4b24}}{2}$. Therefore, $\displaystyle2=(2)\frac{2+a\color{red}{+}\sqrt{a^2+4a+4b24}}{2}+a\Longrightarrow0=a^2+4a+4b24$ When the slope is $6$ (i.e., line $BC$), then $\displaystyle{x}_{BC}=\frac{a6\color{red}{}\sqrt{a^212a+4b+8}}{2}$. Therefore, $\displaystyle6=(2)\frac{a6\color{red}{}\sqrt{a^212a+4b+8}}{2}+a\Longrightarrow0=a^212a+4b+8$ 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. 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. 

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