(a)
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
(a)
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 direct... show full transcript
Worked Solution & Example Answer:(a)
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 \( y - y_0 = \frac{a^2}{b^2 x_0} (x - x_0) \)
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Answer
To find the equation of the normal at point ( P(x_0, y_0) ), we first need the slope of the tangent line. The implicit differentiation of the ellipse equation yields the slope at ( P ). After obtaining the slope ( m ), the slope of the normal will be ( -\frac{1}{m} ). Thus, the equation of the normal can be given by the point-slope form:
[ y - y_0 = -\frac{1}{m}(x - x_0) ]
Substituting the expression for ( m ) will lead you to the final equation.
Step 2
Show that N has coordinates \( (e^2 x_0, 0) \)
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Answer
Since the normal intersects the x-axis, we set ( y = 0 ) in the equation of the normal. Rearranging the resulting equation will allow us to solve for ( x ), yielding the x-coordinate of point ( N ) as ( e^2 x_0 ).
Step 3
Show that \( \frac{PS}{PS'} = \frac{NS}{NS'} \)
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Answer
Using the focus-directrix definition, we have:
[ PS = \text{distance from } P \text{ to } S ] and [ PS' = \text{distance from } P \text{ to } S' ]. Similarly for ( NS ) and ( NS' ). The ratio can be derived from the distances using the coordinates of points and the property of the ellipse.
Step 4
By applying the sine rule to \( \triangle LSP'N \) and to \( \triangle NPS \), show that \( \alpha = \beta \)
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Answer
By applying the sine rule:
[ \frac{LP}{LN} = \frac{SP'}{SN} ] for triangle ( LSP'N ) and
[ \frac{NP'}{NS} = \frac{SP}{LN} ] for triangle ( NPS ).
Equate both ratios to show that ( \alpha = \beta ) considering the angles in the triangles.