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)