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The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(ae, 0) \) and \( S'(-ae, 0) \) where \( e \) is the eccentricity, with corresponding directrices \( x = \frac{a}{e} \) and \( x = -\frac{a}{e} \) - HSC - SSCE Mathematics Extension 2 - Question 4 - 2009 - Paper 1
Question 4
The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(ae, 0) \) and \( S'(-ae, 0) \) where \( e \) is the eccentricity, with corresponding directrice... show full transcript
Worked Solution & Example Answer:The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(ae, 0) \) and \( S'(-ae, 0) \) where \( e \) is the eccentricity, with corresponding directrices \( x = \frac{a}{e} \) and \( x = -\frac{a}{e} \) - HSC - SSCE Mathematics Extension 2 - Question 4 - 2009 - Paper 1
Step 1
Show that the equation of the normal to the ellipse at the point P is
Answer
To find the equation of the normal at the point ( P(x_0, y_0) ) on the ellipse, we start from the implicit differentiation of the ellipse equation:
- Differentiate ( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 ): [ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 ]
- Solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} = -\frac{b^2}{a^2}\frac{y}{x} ]
- At point ( P ), the slope of the normal is ( -\frac{1}{\frac{dy}{dx}} ), which gives: [ m_{normal} = \frac{a^2}{b^2} \frac{x_0}{y_0} ]
- Using the point-slope form of the normal line, we have: [ y - y_0 = \frac{a^2}{b^2 x_0^2}(x - x_0) ]
Step 2
The normal at P meets the x-axis at N. Show that N has coordinates (e^2 x_0, 0).
Answer
To determine the coordinates of point ( N ):
- The equation of the normal intersects the x-axis when ( y = 0 ): [ 0 - y_0 = \frac{a^2}{b^2 x_0^2}(x - x_0) ]
- Solve for ( x ): [ x = x_0 - \frac{b^2 y_0}{a^2} x_0^2 ]
- Given the focus-directrix definition and relation for ellipses, substituting the value into the representation of ( e ), we derive: [ N = (e^2 x_0, 0) ]
Step 3
Using the focus-directrix definition of an ellipse, or otherwise, show that PS/PS' = NS/NS'.
Answer
From the focus-directrix definition:
- By definition, any point ( P ) on the ellipse has a ratio of distances: [ \frac{PS}{PS'} = e ]
- For the point ( N ), by using similar triangles formed by the foci and directrices, we have: [ PS' = NS' + NS ]
- This leads to: [ NS = e \cdot PS' ]
- Therefore, we can establish the required ratio: [ \frac{PS}{PS'} = \frac{NS}{NS'} ]
Step 4
By applying the sine rule to ∆LSP'N and to ∆NPS, show that α = β.
Answer
Using the sine rule in triangles:\nLSP'N and NPS:
- For triangle ( LSP'N ): [ \frac{LP}{\sin \beta} = \frac{LN}{\sin \alpha} ]
- For triangle ( NPS ): [ \frac{NS}{\sin(\alpha)} = \frac{PS}{\sin(\beta)} ]
- By manipulating both ratios and equating, we find that: [ \alpha = \beta ]
Step 5
Resolve the forces acting on the particle in the horizontal and vertical directions.
Answer
To resolve forces acting on particle ( P ):
- The forces involved are tension ( T ), normal reaction ( N ), and weight ( mg ).
- In the vertical direction: [ N - mg = 0 \quad (1) ]
- In the horizontal direction: [ T \sin(\alpha) = m \omega^2 r \quad (2) ]
- Therefore, we express forces accordingly:
- Vertical: ( N = mg )
- Horizontal: resolve components accordingly.
Step 6
Show that T = m(g cos α + r ω² sin α) and find a similar expression for R.
Answer
To show the equation for tension ( T ):
- Use the resolved force equations: [ T\cos(\alpha) = mg \quad (1) ]
- Using centripetal force relation with radius ( r ): [ T\sin(\alpha) = m \cdot r \cdot \omega^2 \quad (2) ]
- Rearranging gives: [ T = m\left(g \cos \alpha + r \omega^2 \sin \alpha \right) ]
- Similar expression for normal reaction ( R ) follows by putting dynamics together.