Submitted by jtedds on Sun, 04/30/2017 - 20:54
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For the first part of the question, I formed two equations $y =\frac{a\sin(\theta)}{b\cos(\theta)}x$ and $y - b\ sin(\theta) = \frac{b\sin(\theta)}{ea + a\cos(\theta)}(x-a\cos(\theta))$
Solving those obtains $y = \frac{bsin(\theta)}{1 + e\cos(\theta)}$ and $x=\frac{b^2\cos(\theta)}{a + ae\cos(\theta)}
I assumed for the second part that I could just sub both $x$ and $y$ straight into $(x+ea)^2 + y^2 = a^2$ but the fractions I end up manipulating seem overly complex so I was wondering if there was an easier way to go about it
The latex in the middle
The latex in the middle should read $x=\frac{b^2\cos(\theta)}{a + aecos(\theta)}$
Looking back at my solution
Looking back at my solution to this question, that I wrote when I was preparing for STEP, I just subbed those expressions for $x$ and $y$ into $(x+ea)^2+y^2$ and showed that it equals $a^2$. The algebra wasn't too bad though; it took me around 2/3 of a page to get the answer. Bear in mind that the algebra in STEP is sometimes going to be quite a bit more involved than in A-Level questions (though if you're doing STEP III questions now, you've probably already realised that). If it helps, I can tell you that those expressions agree with what I have.
Thank you! I had another
Thank you! I had another crack at it but simplified my x coordinate to $x= \frac{a\cos\theta + ea}{e\cos\theta + 1}$ and that seemed to do the trick