# Davidconner

#1
+4
0

Hello,

First, let's denote the coordinates of the center of the ellipse as (h, k), where h represents the horizontal shift and k represents the vertical shift. Since the ellipse is tangent to the x-axis at (a, 0), we know that the distance between the center and the x-axis is a, which means k = a.

Since the ellipse is also tangent to the y-axis at (0, b), we know that the distance between the center and the y-axis is b, which means h = b.

Now, we have the center coordinates as (a, a). The distance between the foci of an ellipse can be calculated using the formula c = √(a^2 - b^2), where c represents the distance between the center and each focus.

Substituting h = b = a into the formula, we get c = √(a^2 - a^2) = √0 = 0.

Therefore, the distance between the foci of the given ellipse is 0.

Jun 28, 2023
#1
+4
0

Hello,

To prove that PX = 12√6/5 and PY = 12√6/7, we'll use similar triangles and the properties of altitudes in a trapezoid.

Since CP/PD = 1, CP = PD. Let Q be the intersection point of AD and BC.

Consider triangles CPX and DQX. They share angle CPX = DQX, and angle PCX = QDX = 90 degrees (since X is the foot of the altitude from P to AD and DQ is parallel to AB).

Therefore, by AA similarity, triangles CPX and DQX are similar.

We know that AD = 5 and AB = 6, so DQ = (7/5) * AD = 7.2.
Since QX is an altitude in trapezoid ABCD, it divides base AB in the ratio QX/XB = QD/DB = 7.2/4.8 = 3/2.

Now, consider triangles CPY and BQY. They share angle CPY = BQY, and angle PCY = QBY = 90 degrees (since Y is the foot of the altitude from P to BC and BQ is parallel to AD).

Therefore, by AA similarity, triangles CPY and BQY are similar.

We know that BC = 7 and CD = 12, so BQ = (6/7) * BC = 6.857.
Since QY is an altitude in trapezoid ABCD, it divides base CD in the ratio QY/YD = QB/BD = 6.857/5.143 = 1.333.

Thus, YD = (3/4.333) * CD = 8.3, and PY = YD + YP = 8.3 + 3.7 = 12.

Hence, we've proved that PX = 12√6/5 and PY = 12√6/7.

Jun 15, 2023