Maximizing Perimeter of Triangle PDE on a Parabola and Finding Coordinates of N for Rhombus Formation

Question

Given a parabola in the Cartesian plane defined by the equation ( $y = -\frac{1}{2}x^2 + \frac{3}{2}x + 2 $), it intersects the x-axis at points A and B, and the y-axis at point C. Consider a point P on the parabola above line BC. A perpendicular line from P to BC meets at point D, and a line parallel to the y-axis through P intersects line BC at point E. It is known that the perimeter of triangle PDE is maximized at ( $\frac{6\sqrt{5}}{5} + 2$ ) when P is at ( $(2, 3)$ ).

The parabola is then translated along the ray CB by ( $\sqrt{5}$ ) units, placing point M on the axis of symmetry of the new parabola. We are to find all coordinates of point N such that the quadrilateral formed by points A, P, M, and N is a rhombus.

Key Points:

• How can we calculate all possible positions for point N based on these conditions?

Attempt: I calculated the intersection points of the parabola with the axes and found the coordinates for points A and B by solving the quadratic equation from the parabola's formula. I derived the perimeter condition for triangle PDE and established point P's position at ( $(2, 3)$ ).

For translating the parabola, I used the vector direction from point C to B to determine the translation vector. However, I need help understanding how to correctly apply the rhombus condition to find all possible coordinates for point N.

Question: Can anyone provide insight or methods to determine the coordinates of point N such that APNM forms a rhombus? What are the necessary geometric or algebraic steps to resolve this part of the problem?

Answer 1

Diagonal AN must be perpendicular on diagonal MP. We need to find the coordinates of F, the mid point of MP and AN. We have:

$A(-1, 0)$, $B(4, 0)$, $C((0, 2)$, $P(2, 3)$

1-Equation of BC:$\frac {x-0}{4-0}=\frac{y-2}{0-2}\rightarrow y=-\frac12 x+2$

2- equation of PM: $m=-\frac12\rightarrow y-3=-\frac12 (x-2)\rightarrow y=-\frac 12 x +4$

3-Equation of AN:$m=\frac{-1}{-\frac 12}=2\rightarrow y=2x+2$

4-(2) with (3) give $F(\frac 45, \frac {18}5)$

5-$x_F=\frac{x_A+x_N}2\Rightarrow x_N=2x_F-x_A=2(\frac45)-(-1)=\frac {13}5$

6-$y_F=\frac{y_A+y_N}2\Rightarrow y_N=2y_F-y_A=2(\frac{18}5)-0=\frac {36}5$

$N(\frac{13}5, \frac{36}5)$

I think this is the only possible point.Note that $CE=\sqrt{2^2+1^2}=\sqrt 5$

The equation of new parabola is:

$-\frac1{1.04}y=(x-2.53)(x+3.33)$