I have a test on Monday, Help me out, please.

Accepted Solution

Answer:[tex]\frac{1}{\lambda^2}\sqrt{1-\lambda^2+2\lambda^3+2\lambda^4}[/tex] is the length of ABStep-by-step explanation: We want to find the area enclosed my connecting the three points  [tex]O(0,0)[/tex], [tex]A(-1,1)[/tex], and [tex]B(\frac{1}{\lambda},\frac{1}{\lambda^2})[/tex] where [tex]\lambda[/tex] is positive.We are going to use AB has the base length of this triangle.We are going to use OP has the height of this triangle.P will be the intersection point of AB and OP such that OP is perpendicular to AB.  Once we find the base and the height, we will be able to find the area of triangle using the following formula:[tex]\frac{1}{2}(\text{base})(\text{height})[/tex].I will find [tex]\lambda[/tex] later so that AB is at maximum height.Let's begin.I could only fit the question I labeled Problem 2 above:Length of ABSo to compute length AB I'm going to use the distance formula:[tex]\sqrt{(\frac{1}{\lambda}-(-1))^2+(\frac{1}{\lambda^2}-1)^2}[/tex][tex]\sqrt{(\frac{1}{\lambda}+1)^2+(\frac{1}{\lambda^2}-1)^2}[/tex][tex]\sqrt{\frac{1}{\lambda^2}+\frac{2}{\lambda}+1+\frac{1}{\lambda^4}-\frac{2}{\lambda^2}+1}[/tex][tex]\sqrt{\frac{1}{\lambda^4}-\frac{1}{\lambda^2}+\frac{2}{\lambda}+2}[/tex]Factor out the [tex]\sqrt{\frac{1}{\lambda^4}}[/tex] since this square root contains a perfect square:[tex]\frac{1}{\lambda^2}\sqrt{1-\lambda^2+2\lambda^3+2\lambda^4}[/tex].This last line is as far as I'm going to simplify the distance between points A and B. [tex]\frac{1}{\lambda^2}\sqrt{1-\lambda^2+2\lambda^3+2\lambda^4}[/tex] is the length of AB.Side note: I did use the formula [tex](a+b)^2=a^2+2ab+b^2[/tex] to help me expand the squared differences/sums.