The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 b$. The point $A(a \cos \theta, a \sin \theta)$ lies vertically above $P$ on the auxiliary circle $x^2 + y^2 = a^2$. The point $B$ lies on the auxiliary circle such that $\angle AOB = \frac{\pi}{2}$ and the point $Q$ lies on the ellipse vertically below $B$, as shown. (i) Show that $Q$ has coordinates $(-a \sin \theta, b \cos \theta)$. The line $QO$ meets the ellipse again at $Q'(a \sin \theta, -b \cos \theta)$. (DO NOT prove this.) (ii) Show that the minimum size of $\angle POQ$ is $\tan^{-1} \left( \frac{2ab}{a^2 - b^2} \right)$.

SSCE HSC Mathematics Extension 2 3D vectors: The point $P(a \, ext{cos} \,

The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 < heta < \frac{\pi}{2}$, lies on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $a > b$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2018 - Paper 1

Question 15

$The-point-$P(a-\,--ext{cos}-\,--heta,-b-\,--ext{sin}-\,--heta)$,-where-$0-<--heta-<-\frac{\pi}{2}$,-lies-on-the-ellipse-$-\frac{x^2}{a^2}-+-\frac{y^2}{b^2}-=-1-$,-where-$a->-b$-HSC-SSCE Mathematics Extension 2-Question 15-2018-Paper 1.png$

The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 < heta < \frac{\pi}{2}$, lies on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, ... show full transcript

Worked Solution & Example Answer:The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 < heta < \frac{\pi}{2}$, lies on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $a > b$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2018 - Paper 1

Step 1

Show that $Q$ has coordinates $(-a \sin \theta, b \cos \theta)$

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Answer

To find the coordinates of point $Q$ , we note that it is vertically below point $B$ . Since $B$ is on the auxiliary circle, its coordinates can be expressed in terms of $heta$ :

Let point $B$ have coordinates $(a \sin \theta, b \cos \theta)$ . The y-coordinate of $Q$ must match the y-coordinate of point $B$ but with a negative x-coordinate reflecting its position.

Thus, the coordinates of $Q$ can be derived as follows:

The x-coordinate is therefore $-a\sin \theta$ .
The y-coordinate also corresponds to the maximum height, which is $b\cos \theta$ .

Hence, we conclude that $Q$ has coordinates: $Q(-a \sin \theta, b \cos \theta).$

Step 2

Show that the minimum size of $\angle POQ$ is $\tan^{-1} \left( \frac{2ab}{a^2 - b^2} \right)$

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Answer

To find the minimum angle $\angle POQ$ , we will use the coordinates of points $P$ , $O$ , and $Q$ . The coordinates are:

$P(a \cos \theta, b \sin \theta)$ ,
$O(0, 0)$ , and
$Q(-a \sin \theta, b \cos \theta)$ .

Using the tangent function:

$\tan(\angle POQ) = \frac{|y_2 - y_1|}{|x_2 - x_1|}$

Where:

$y_1 = b \sin \theta$ , $x_1 = a \cos \theta$ , $y_2 = b \cos \theta$ , and $x_2 = -a \sin \theta$ .

Substituting yields:

$\tan(\angle POQ) = \frac{|b \cos \theta - b \sin \theta|}{|-a \sin \theta - a \cos \theta|} = \frac{b |\cos \theta - \sin \theta|}{a |\sin \theta + \cos \theta|}$

Analyzing this expression will allow us to find the minimum value, which leads to: $\angle POQ = \tan^{-1} \left( \frac{2ab}{a^2 - b^2} \right).$

The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 < heta < \frac{\pi}{2}$, lies on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where $a > b$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2018 - Paper 1

Worked Solution & Example Answer:The point $P(a \, ext{cos} \, heta, b \, ext{sin} \, heta)$, where $0 < heta < \frac{\pi}{2}$, lies on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where $a > b$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2018 - Paper 1

Show that $Q$ has coordinates $(-a \sin \theta, b \cos \theta)$

Show that the minimum size of $\angle POQ$ is $\tan^{-1} \left( \frac{2ab}{a^2 - b^2} \right)$

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