# Proof of an ellipse

• Mar 30th 2010, 03:35 AM
cozza
Proof of an ellipse
The question I am stuck on is (iii):

This part concerns the conic with equation
$\displaystyle 3x^2+5y^2=75$

(i) By rearranging the equation of the conic, classify it as an ellipse, parabola or hyperbola in standard position, and sketch the curve.

I have re-arranged the equation and discovered it is an ellipse in standard position:
$\displaystyle (x^2)/5^3 + (y^2)/(\sqrt{15})^2 = 1$
a=5, b=$\displaystyle \sqrt{15}$
The points of the ellipse are (-5,0), (0, $\displaystyle \sqrt{15}$), (5,0) and (0, $\displaystyle -\sqrt{15}$)

(ii) Find exact values for the eccentricity, foci and directrices of this conic, and mark the foci and directrices on your sketch.

eccentricity (e) = $\displaystyle \sqrt{1-b^2/a^2}$= $\displaystyle \sqrt{10}/5$
foci = +-ae = $\displaystyle (-\sqrt{10}, 0)$ and $\displaystyle (\sqrt{10}, 0)$
directrices = +-a/e = $\displaystyle +-25/\sqrt{10}$

(iii) Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.

point, P where the conic intersect the x-axis ... etc. (-5,0)
focus with negative x co-ordinate = $\displaystyle (-\sqrt{10},0)$
corresponding directrix = $\displaystyle -25/\sqrt{10}$
eccentricity = $\displaystyle \sqrt{10}/5$

I know that PF = a-ae = $\displaystyle -5+\sqrt{10}$
and ePd = e(a-ae)/e = $\displaystyle \sqrt{10}/5\times(-5+\sqrt{10}/\sqrt{10}/5$
= $\displaystyle -5+\sqrt{10}$

so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.

I hope what I have done so far is correct? Thanks again for your help. It is most appreciated (Happy)
• Mar 30th 2010, 04:02 AM
tonio
Quote:

Originally Posted by cozza
The question I am stuck on is (iii):

This part concerns the conic with equation
$\displaystyle 3x^2+5y^2=75$

(i) By rearranging the equation of the conic, classify it as an ellipse, parabola or hyperbola in standard position, and sketch the curve.

I have re-arranged the equation and discovered it is an ellipse in standard position:
$\displaystyle (x^2)/5^3 + (y^2)/(\sqrt{15})^2 = 1$
a=5, b=$\displaystyle \sqrt{15}$
The points of the ellipse are (-5,0), (0, $\displaystyle \sqrt{15}$), (5,0) and (0, $\displaystyle -\sqrt{15}$)

Surely you meant "the points where the ellipse meets the axis are..."

(ii) Find exact values for the eccentricity, foci and directrices of this conic, and mark the foci and directrices on your sketch.

eccentricity (e) = $\displaystyle \sqrt{1-b^2/a^2}$= $\displaystyle \sqrt{10}/5$
foci = +-ae = $\displaystyle (-\sqrt{10}, 0)$ and $\displaystyle (\sqrt{10}, 0)$
directrices = +-a/e = $\displaystyle +-25/\sqrt{10}$

(iii) Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.

point, P where the conic intersect the x-axis ... etc. (-5,0)
focus with negative x co-ordinate = $\displaystyle (-\sqrt{10},0)$
corresponding directrix = $\displaystyle -25/\sqrt{10}$
eccentricity = $\displaystyle \sqrt{10}/5$

I know that PF = a-ae = $\displaystyle -5+\sqrt{10}$
and ePd = e(a-ae)/e = $\displaystyle \sqrt{10}/5\times(-5+\sqrt{10}/\sqrt{10}/5$
= $\displaystyle -5+\sqrt{10}$

so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.

Any function intersects the x-axis when y=0, and the y-axis when x=0...

Tonio

I hope what I have done so far is correct? Thanks again for your help. It is most appreciated (Happy)

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• Mar 30th 2010, 04:05 AM
sa-ri-ga-ma
so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.

The other point of intersection is (5, 0).
For this point, distances to the directrix with negative x co-ordinate is PF = (a + ae) , and PD is (2a + a/e)
• Mar 30th 2010, 04:15 AM
cozza
Quote:

Originally Posted by sa-ri-ga-ma
so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.

Quote:

Originally Posted by sa-ri-ga-ma

The other point of intersection is (5, 0).
For this point, PF = (a + ae) , and PD is (2a + a/e)

Thank you for the quick reply. I understand that the other point of intersection is (5,0), but the question specifically says:

Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.

Surely there is only one point where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity?

Am I missing something here? At the point (5,0) the f has a positive x-coordinate, as does the corresponding directrix, doesn't it?

Thanks again :)
• Mar 30th 2010, 04:24 AM
sa-ri-ga-ma
At the point (5,0) the f has a positive x-coordinate, as does the corresponding directrix, doesn't it?

Here you have to take the distance of the point P (5,0) from negative focus and the corresponding diretrix. Then prove the relation PF = e*PD.