7 Given that \(x = 2 \sec \theta\) and \(y = 3 \tan \theta\), show that \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1\).
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Question 7:
\(\sec\theta = x/2\), \(\tan\theta = y/3\)
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Guidance
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Guidance
\(\sec^2\theta = 1 + \tan^2\theta\) M1
\(\sec^2\theta = 1 + \tan^2\theta\) used (oe, e.g. converting to sines and cosines and using \(\cos^2\theta + \sin^2\theta = 1\))
\(\frac{x^2}{4} = 1 + \frac{y^2}{9}\) M1
Eliminating \(\theta\) (or for \(x\) and \(y\))
\(\Rightarrow \frac{x^2}{4} - \frac{y^2}{9} = 1\)* E1
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OR \(x^2/4 - y^2/9 = 4\sec^2\theta/4 - 9\tan^2\theta/9 = \sec^2\theta - \tan^2\theta = 1\)[3]
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## Question 7:
$\sec\theta = x/2$, $\tan\theta = y/3$
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sec^2\theta = 1 + \tan^2\theta$ | M1 | $\sec^2\theta = 1 + \tan^2\theta$ used (oe, e.g. converting to sines and cosines and using $\cos^2\theta + \sin^2\theta = 1$) |
| $\frac{x^2}{4} = 1 + \frac{y^2}{9}$ | M1 | Eliminating $\theta$ (or for $x$ and $y$) |
| $\Rightarrow \frac{x^2}{4} - \frac{y^2}{9} = 1$* | E1 | www |
| **OR** $x^2/4 - y^2/9 = 4\sec^2\theta/4 - 9\tan^2\theta/9 = \sec^2\theta - \tan^2\theta = 1$ | | |
| | **[3]** | |
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7 Given that $x = 2 \sec \theta$ and $y = 3 \tan \theta$, show that $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1$.
\hfill \mbox{\textit{OCR MEI C4 Q7 [3]}}