+0

# ellipses

0
87
1

For $$0 < k < 6,$$ the graphs of $$\frac{(x - k)^2}{9} + y^2 = 1$$ and $$\frac{x^2}{9} + y^2 = 1$$ intersect at $$A$$ and $$C,$$ and have $$x$$-intercepts at $$B$$and $$D$$ respectivly. Compute the value of $$k$$ for which $$ABCD$$ is a square.

Dec 4, 2018
edited by Guest  Dec 4, 2018

#1
+5073
0

You'll have to work through much of the algebra here yourself.

$$\text{you have two ellipses of the same size, one centered at } (k,0), \text{ the other at }(0,0)$$

$$\text{Looking at the geometry it must be that }B \text{ is the leftmost x intercept of }\dfrac{(x-k)^2}{9}+y^2=1\\ \text{and }D \text{ is the rightmost x intercept of }\dfrac{x^2}{9}+y^2=1$$

$$B = (k-3,0)\\ D=(3,0)$$

$$\text{The }x \text{ coordinates of the intersections are given by}\\ (x-k)^2=x^2\\ -2kx+k^2 = 0\\ x = \dfrac{k}{2}\\ y = \pm \sqrt{1-\dfrac{k^2}{36}}$$

$$\text{the quadrilateral formed by joining these points is a rhombus, so we solve for }k\\ \text{that makes the corner angle 90 degrees}\\AD = A-D = \left(\dfrac k 2,\sqrt{1-\dfrac{k^2}{36}}\right)-(3,0) = \left(\dfrac k 2 - 3,\sqrt{1-\dfrac{k^2}{36}}\right)\\ CD=C-D = \left(\dfrac k 2,-\sqrt{1-\dfrac{k^2}{36}}\right)-(3,0) = \left(\dfrac k 2 - 3,-\sqrt{1-\dfrac{k^2}{36}}\right)\\ \text{the angle between }AD \text{ and }CD \text{ being 90 degrees} \Rightarrow AD \cdot CD = 0$$

$$AD \cdot CD = \dfrac{5 k^2}{18}-3 k+8=0\\$$

$$k = \dfrac{24}{5} \text{ or the size zero square when }k=6$$

.
Dec 4, 2018
edited by Rom  Dec 4, 2018
edited by Rom  Dec 4, 2018
edited by Rom  Dec 4, 2018