Question

Hi, could anyone help me solve the following question?
Please find attached.

Answer 1

If you can use derivatives, then one possible way could be the following,
\[y_1=\sqrt{1-x^2}\]for the circunference, and \[y_2=\frac{9}{16}\sqrt{\frac{16}{9}-x^2}\]for the ellipse. As a tangent line to both figures should have the same slope, then, there must be x that makes both derivatives equal. So, calculate the derivatives and then solve\[y'_1=y'_2\]. You will find two values for x (plus and minus sign), and then, two values for y.
But remember that there must be 4 lines (one for each quadrant).

Answer 2

okay. But it seems your y2 in terms of x is not correct

Answer 3

You're right. It should be,
\[y_2=\frac{3}{4}\sqrt{\frac{16}{9}-x^2}\]

Answer 4

By the way, so there are two roots for x. After finding out x, I substitute the values of x into either y1' or y2' to get the answer?

Answer 5

The x is quite a complicated number

Answer 6

You should substitute x in both y1' and y2'. This will give two of the four lines that are tanget to the figures of the problem.

Answer 7

actually what about y1' = -y2'? Just now you simply take the square roots without considering the negative sign. There are four possibilities.

Answer 8

oh. my apology. two. y1'=y2' or y1' = -y2'

Answer 9

Yes, I think that if you want to solve this equations you wrote, you must to calculate the square of both sides (to avoid square roots). So the sign will not help you.

Answer 10

Did you get x^2 = (3367)/(1575). ? this answer is....

Answer 11

I obtain x=5/3.

Answer 12

how? Did you use the new y2?

Answer 13

wait. your y2 is not correct again

Answer 14

I'll check it again ;).

Answer 15

inside the square root, it should be 1-（9/16）x^2

Answer 16

well, I was thinking of another approach. Use implicit differentiation.
For x^2 + y^2 = 1, the dy/dx = -x/y
For the ellipse, at point (x0,y0), the equation of tangent is x0x/(4/3)^2 + y0y/ (3/4)^2 = 1
But how to combine these two equations of the tangent?

Answer 17

Well, I checked it and I get,
\[x=\frac{\sqrt{481}}{15}\]

Answer 18

The derivatives I used,
\[y'_1=\frac{-x}{\sqrt{1-x^2}}\\
y'_2=-\frac{27}{64}\frac{x}{\sqrt{1-9x^2/16}}\]

Answer 19

I'll check it. thanks for the help. Really appreciate it!

Answer 20

You're welcome. Only to complete a little, once you get the x coordinate (plus and minus), it would be easier to substitute each value in the original equations (circunference and ellipse). For each one you will find 2 values of y (plus and minus), that let you find the four lines of the form indicated by,
\[y-f(x_0)=f'(x_0)(x-x_0)\]

Answer 21

The value I give you as a solution it has no sense. Seems we have to work out a little more this problem.

Answer 22

Also, you cannot use the method of derivatives in the form I wrote it.

Answer 23

Well, I found a way that it can be usefull. First, you know that the tangent should touch each figure in one and only one point. The equation for this tanget should be of the form,
\[y=mx+n\]
Now you put this equation in the equations of both figures, and solve for m and n,
\[x^2+y^2=1⇒n^2=m^2+1\\
(x/a)^2+(y/b)^2=1⇒n^2=a^2m^2+b^2\]

Solving the system obtained for m and n, we have,
\[m=\pm\sqrt{\frac{1−b^2}{a^2−1}}\\
n=±\sqrt{\frac{a^2−b^2}{a^2−1}}\]
Where a=4/9 and b=3/4.
This will give you the four equations,
\[y=3x/4+5/4\\
y=3x/4−5/4\\
y=−3x/4+5/4\\
y=−3x/4−5/4\]

And the plot I attach.

Answer 24

Hi, I think you have found the answer. Thank you! But how did you at first sub y=mx+n into the two equations?

Answer 25

\[x^2+y^2=1\Rightarrow x^2+(mx+n)^2=1\Rightarrow x^2+m^2x^2+n^2+2mnx=1\\
x^2(1+m^2)+2mnx+n^2-1=0\]As the tangent line must touch in one point and only in one point the discriminant of the second grade equation must be zero,
\[(2mn)^2-4\cdot(1+m^2)(n^2-1)=0\Rightarrow n^2=m^2+1\]

Answer 26

And the same for the ellipse.

Answer 27

I see. Well, that's a nice approach

Answer 28

Have you tried using implicit differentiation?

Answer 29

Yes, but I don't see how to continue the problem after finding dy/dx.