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(a) The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(a e, 0) \) and \( S'(-a e, 0) \) 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

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(a)-The-ellipse-\(-\frac{x^2}{a^2}-+-\frac{y^2}{b^2}-=-1-\)-has-foci-\(-S(a-e,-0)-\)-and-\(-S'(-a-e,-0)-\)-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.png

(a) The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(a e, 0) \) and \( S'(-a e, 0) \) where \( e \) is the eccentricity, with corresponding dire... show full transcript

Worked Solution & Example Answer:(a) The ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) has foci \( S(a e, 0) \) and \( S'(-a e, 0) \) 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 \( y - y_0 = \frac{a^2}{b^2 y_0}(x - x_0) \)

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Answer

To find the normal at point ( P(x_0, y_0) ), we need:

  1. The derivative of the ellipse equation, which is given by implicit differentiation:

    dydx=−b2xa2y\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}

    By substituting ( x = x_0 ) and ( y = y_0 ), we get the slope of the tangent line. The negative reciprocal of this slope will give us the slope of the normal line.

  2. The normal line through point ( P ) can be written in point-slope form:

    y−y0=m(x−x0)y - y_0 = m (x - x_0)

    where ( m ) is the negative reciprocal of the tangent slope. On further simplification, we find that:

    y−y0=a2b2y0(x−x0).y - y_0 = \frac{a^2}{b^2 y_0} (x - x_0).

Step 2

Show that N has coordinates \( (e^2 x_0, 0) \)

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Answer

The normal meets the x-axis at point ( N ). At this intersection, ( y = 0 ). Using the normal equation derived in part (i):

0−y0=a2b2y0(x−x0)0 - y_0 = \frac{a^2}{b^2y_0} (x - x_0)

Solving for ( x ) gives:

x=x0+b2y0a2y0.x = x_0 + \frac{b^2y_0}{a^2}y_0.

We know ( e = \frac{c}{a} ) and that ( c = ae ). Thus:

Substitute into the coordinate system to find that:

N=(e2x0,0).N = (e^2 x_0, 0).

Step 3

Show that \( \frac{PS}{PS'} = \frac{NS}{NS'} \)

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Answer

By the definition of an ellipse, we know:

  • The distances from point ( P ) to the foci ( S ) and ( S' ): ( PS ) and ( PS' ).
  • The distances from point ( N ) to the same foci: ( NS ) and ( NS' ).

It follows that:

PS+PS′=constant,PS + PS' = \text{constant}, which implies: PSPS′=NSNS′.\frac{PS}{PS'} = \frac{NS}{NS'}.

Step 4

By applying the sine rule to \( \triangle Z'S'PN \) and to \( \triangle NPS \), show that \( \alpha = \beta \)

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Answer

Applying the sine rule to triangles gives:

For ( \triangle Z'S'PN ): Z′S′sin⁡α=PNsin⁡Z′PN\frac{Z'S'}{\sin \alpha} = \frac{PN}{\sin Z'PN}

For ( \triangle NPS ): NSsin⁡β=PSsin⁡NPS\frac{NS}{\sin \beta} = \frac{PS}{\sin NPS}

Setting both expressions for ( PS ) and using the equality:

Hence, we conclude: ( \alpha = \beta ).

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