The point P $(a\cos\theta, b\sin\theta)$, 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 $\tan^{-1}(\frac{2ab}{a^2 - b^2})$.

SSCE HSC Mathematics Extension 2 Geometry proofs using vectors: The point P $(a\cos\theta, b\sin

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

Worked Solution & Example Answer:The point P $(a\cos\theta, b\sin\theta)$, where $0 < \theta < \frac{\pi}{2}$, 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 $(-a\sin\theta, b\cos\theta)$

96%

114 rated

Answer

To find the coordinates of point Q, we note that it lies directly below point B. Since B is on the auxiliary circle, we have:

The coordinates of B can be derived from the angle AOB. Given that ( \angle AOB = \frac{\pi}{2} ), the coordinates are derived using the properties of the auxiliary circle:

B has the same x-coordinate as A, which is $a\cos\theta$ . Hence, the x-coordinate of Q will be:

[ x_Q = -a\sin\theta ]
The y-coordinate can be inferred from the fact that Q lies on the ellipse, translating into:

[ \frac{x_Q^2}{a^2} + \frac{y_Q^2}{b^2} = 1 ]

Substituting the x-coordinate:

[ \frac{(-a\sin\theta)^2}{a^2} + \frac{y_Q^2}{b^2} = 1 ]

This simplifies to:

[ \sin^2\theta + \frac{y_Q^2}{b^2} = 1 ]

Rearranging gives us:

[ \frac{y_Q^2}{b^2} = 1 - \sin^2\theta = \cos^2\theta ]

Thus, yielding:

[ y_Q = b\cos\theta ]

Therefore, the coordinates for Q are ( (-a\sin\theta, b\cos\theta) ).

Step 2

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

99%

104 rated

Answer

To find the minimum angle ( \angle POQ ), we will use the tangent definition in relation to the coordinates we derived:

The coordinates for P are $(a\cos\theta, b\sin\theta)$ and for Q are $(-a\sin\theta, b\cos\theta)$ .
The slope of line OP is:

[ m_{OP} = \frac{b\sin\theta - 0}{a\cos\theta - 0} = \frac{b\sin\theta}{a\cos\theta} ]

The slope of line OQ is:

[ m_{OQ} = \frac{b\cos\theta - 0}{-a\sin\theta - 0} = -\frac{b\cos\theta}{a\sin\theta} ]
The angle ( \angle POQ ) can be computed using the tangent of the angle difference formula:

[ \tan(\angle POQ) = \left| \frac{m_{OP} - m_{OQ}}{1 + m_{OP} m_{OQ}} \right| ]

Substituting the expressions for $m_{OP}$ and $m_{OQ}$ gives:

[ \tan(\angle POQ) = \left| \frac{\frac{b\sin\theta}{a\cos\theta} + \frac{b\cos\theta}{a\sin\theta}}{1 - \frac{b^2}{a^2}} \right| ]

Simplifying further leads to:

[ \tan(\angle POQ) = \frac{2ab}{a^2 - b^2} ]
Therefore, the minimum angle size of ( \angle POQ ) is:

[ \angle POQ = \tan^{-1}(\frac{2ab}{a^2 - b^2}) ]

The point P $(a\cos\theta, b\sin\theta)$, where $0 < \theta < \frac{\pi}{2}$, 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 $(a\cos\theta, b\sin\theta)$, where $0 < \theta < \frac{\pi}{2}$, 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 $\tan^{-1}(\frac{2ab}{a^2 - b^2})$

Join the SSCE students using SimpleStudy...