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