Alan

You can try the other one using the same technique, bearing in mind that the foci are on the y-axis rather than the x-axis this time.

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Equation of ellipse:  (x/a)^2 + (y/b)^2 = 1,  where a and b are semi-major axes.

Here a = 7 from the fact that the vertex is at (7, 0) so a^2 = 49

There are two foci. One is given as (2, 0) so the other must be at (-2, 0).

The sum of the distance from one focus to the ellipse to the other focus is constant. Here it must be (7 - 2) + (7 + 2) → 14

The y vertex is at (0, b) so we must also have 2*sqrt(b^2 + 2^2) = 14

b^2 + 2^2 = 7^2. or b^^2 = 45

Hence we can write the ellipse equation as x^2/49 + y^2/45 = 1

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Replace x by h + 2. The ratio becomes:  (3(h + 2)^2 - 12)/(h + 2 - 2), or (3h^2 + 12h)/h → 3h + 12

When x → 2 we want h → 0, which means the ratio tends to 12.

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This question was asked and answered here:

http://web2.0calc.com/questions/several-logs-are-stored-in-a-pile-with-22-logs-on-the-bottom-layer-21-on-the-second-layer-20-on-the-third-layer-and-so-on-if-the-top-layer-has-one-log

(4 + x_K)/2 = 4.   Hence x_K = 4

(1 + y_K)/2 = 5.   Hence y_K = 9

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Yes! By expressing it as a fraction.

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The above two answers look different but they are the same to within an arbitrary constant:

sin²x + cos²x = 1

Divide by cos²x

tan²x + 1 = 1/cos²x

or:   tan²x + 1 = sec²x

Try it this way:

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