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

Sep 6, 2020

#1
+9180
+2

Did you mean to put  $$\frac{(x - k)^2}{9} + y^2 = 1$$   and   $$\frac{(x + k)^2}{9} + y^2 = 1$$   ?     If so.....

By finding the x-intercepts and the points of intersection (in terms of k), we can find the slope of line AB, BC, CD, and AD.

Then, once we know a point and the slope of each line, we can find the equations of each line.

And so we can make this graph: https://www.desmos.com/calculator/uozsru2uca

Now we need to find the value of  k  that makes AB perpendicular to BC

(which will also make AB perpendicular to AD, and CD perpendicular to BC, and CD perpendicular to AD)

That is, we need to find the value of  k  that makes the following true:

the slope of  BC  =  the negative reciprocal of the slope of AB

$$-\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)\quad=\quad$$the negative reciprocal of   $$\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)$$

$$-\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)\quad=\quad-\left(\frac{9-3k}{\sqrt{9-k^{2}}}\right)$$

Multiply both sides of the equation by  -1

$$\frac{\sqrt{9-k^{2}}}{9-3k}\quad=\quad\frac{9-3k}{\sqrt{9-k^{2}}}$$

Multiply both sides by  $$\sqrt{9-k^2}$$  and multiply both sides by  $$9-3k$$

$$(\sqrt{9-k^2}\ )^2\quad=\quad(9-3k)^2$$

Simplify the left side and multiply out the right side

$$9-k^2\quad=\quad(9-3k)(9-3k)$$

$$9-k^2\quad=\quad81-54k+9k^2$$

Subtract  9  from both sides and add  k2  to both sides

$$0\quad=\quad72-54k+10k^2$$

Rearrange the terms

$$0\quad=\quad10k^2-54k+72$$

Now we can use the quadratic formula to find  k

$$k\quad=\quad\dfrac{54\pm\sqrt{54^2-4\cdot10\cdot72}}{2\cdot10}$$

$$k\quad=\quad\dfrac{54\pm6}{20}$$

$$k\quad=\quad3\qquad\text{or}\qquad k\quad=\quad\frac{12}{5}$$

If  k = 3  then the points A, B, C, and D are all on the origin. But if  k = 12/5  then ABCD is a regular square.

If you would like more explanation on how to find the slopes of lines BC and AB then please let us know!

Sep 6, 2020
#2
+1041
+1

Hi hectictar

we all missed your presence on the forum!

ilorty  Sep 6, 2020
#3
+9180
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Thanks ilorty

Also I just noticed that the y-axis appears to be to the left of AC in the given picture, but the y-axis is directly on AC in my graph....so maybe I guessed the intended meaning of the question wrong......

hectictar  Sep 6, 2020
#4
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Actually I meant the second equation to be $$\frac{x^2}{9}+y^2 = 1$$ but thank you for the answer anyways.

hulu0104  Sep 7, 2020
#5
+30
0

It was my probem I got the equation wrong.

hulu0104  Sep 7, 2020
#6
+30
0

It's my problem I got the equation wrong, I'm sorry.

hulu0104  Sep 7, 2020
#7
+9180
+1

Can you apply the same method to solve your actual question?

First find the x-intercepts of each graph. (Plug in  0  for  y  and solve for  x )

Then find the points of intersection.  (The solutions to   $$\frac{(x-k)^2}{9}+y^2\ =\ \frac{x^2}{9}+y^2$$   are the points of intersection)

Once you know the coordinates of each point A, B, C, and D, then find the slope of AB and the slope of BC.

(If you calculate the slope of CD and AD you should find that they are the same as AB and BC, respectively)

Here's a new graph:  https://www.desmos.com/calculator/x63mqeac9l

Then find which value of  k  will make the slope of BC = the negative reciprocal of the slope of AB

If you need more help or get stuck let us know!

Sep 7, 2020