WEBVTT
Kind: captions
Language: en
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Hello.
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Welcome to the section, "It's Your Turn On
Ellipses".
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In this exercise, we have to compute the center,
the foci, and the length of the two axes of
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the ellipse, which has this question.
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Good.
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First of all, let us recall which is the equation
of an ellipse with axes which are parallel
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to the coordinate axis.
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The equation in its canonical form.
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OK.
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The equation is the following.
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You have x minus x0 squared over a squared
plus y minus y0 squared over b squared equal
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to 1, where a and b are positive real numbers.
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OK.
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Which is the meaning of x0, y0, a, and b?
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OK.
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The center of this ellipse is the point of
coordinates x0, y0.
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OK.
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Then, the distance between the foci and the
center is given by the square root of the
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module of a squared minus b squared.
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Good.
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And finally, 2a and 2b are the lengths of
the two axes.
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Good.
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Therefore, to solve our exercise, first of
all, we have to transform our equation in
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something which has this shape.
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Good.
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Oh, finally, the last thing.
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Remember that if this denominator is greater
than this denominator, then we will have an
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ellipse which is horizontally oriented, with
the major axis, which is parallel to the x-axis.
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On the other side if this denominator is greater
than this one, then we will have an ellipse,
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which is vertically oriented with the major
axis parallel to the y-axis.
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OK.
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Now let us start with our exercise.
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Good.
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We can re-write this equation in the following
form.
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We have x squared minus 4 times 6 plus 4,
which multiplies y squared plus a 6 times
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y plus 24 equal to 0.
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And now let us complete the squares.
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And then, these two terms here are equal to
what?
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To x minus 2 squared minus 4 plus 4, which
multiplies.
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OK, now, these two terms here are equal to
y plus 3 squared minus 9.
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And then we have our constant term, 24.
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Good.
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And now, do you see it's x minus 2 squared
plus 4 times y plus 3 squared and then you
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have a minus 4 minus 36, which is minus 40,
plus 24, which is minus 16.
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And then moving the minus 16 on the other
side with respect to the equal, I have 16.
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Good.
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But 16, we know that is 4 squared.
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And now we divide by 4 squared on both sides.
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And what do we get?
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We get x minus 2 squared over 4 squared plus--
here we have a 4 here, therefore, dividing
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it by 4 squared, we have at the denominator
we have a 4, which is 2 squared.
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OK?
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And then equal to 1.
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OK.
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We have obtained the equation of our ellipse
in this form.
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And then now we can answer to all the questions
of the exercise.
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Precisely, this is the graph of our ellipse.
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And now what we can say?
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We can immediately say which are the coordinates
of the center because the center is the point
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of coordinates 2 and minus 3.
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Then, as it's clear from the graph, you see--
because this denominator is greater than this
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denominator, our ellipse is horizontally oriented,
and therefore, the foci are on the horizontal
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line passing through the center C. Good.
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And what do we know, we know that the distance
between the foci and the center is equal to,
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what?
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To the square root of the module of the difference
between these two.
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But this is 16, this is 4.
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Then it's the square root of 12.
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OK.
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12 is 4 times 3.
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Therefore, we can rewrite this in a simpler
way, as 2 times the square root of 3.
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Good.
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And knowing that the foci are on the horizontal
line passing through the center, and that
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the distance between the foci and C, and the
centre C is 2 times the square root of 3,
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we immediately get that the focus F1 here
is the point-- OK.
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Because they are on the same horizontal line
as the centre, then the focus F1 will have
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the second coordinate equal to minus 3 and
the first coordinate is 2 minus 2 times the
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square root of 3.
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Such a way, the distance is 2 times the square
root of 3.
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And the other focus is F2 equal to 2 plus
2 times, now, the square root of 3 minus 3.
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Finally.
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OK.
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We have that-- you know-- a here, which is
4 here, is the length of the major axis.
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The major axis, its length is 8.
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And the minor axis, has length 4.
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Good.
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This is the major axis and this is, here,
the minor axis.
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OK.
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Good.
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Thank you very much.