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(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

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(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.png

(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

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Answer

To find the equation of the normal at point ( P(x_0, y_0) ) on the ellipse, we start by differentiating the equation of the ellipse implicitly. The equation is given by:

x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Differentiating both sides with respect to ( x ):

2xa2+2yb2dydx=0\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0

From this, we find the slope of the tangent line at ( P ):

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

Thus, the slope of the normal line, which is the negative reciprocal of the tangent slope, is given by:

m=a2y0b2x0m = \frac{a^2 y_0}{b^2 x_0}

Using the point-slope form of the line equation, the equation of the normal line can be expressed as:

y−y0=m(x−x0)y - y_0 = m (x - x_0)

Substituting for ( m ), we have:

y−y0=a2y0b2x0(x−x0)y - y_0 = \frac{a^2 y_0}{b^2 x_0}(x - x_0)

This simplifies to:

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

Therefore, the equation of the normal is confirmed.

Step 2

The normal at P meets the x-axis at N. Show that N has coordinates (e2x0, 0).

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Answer

To find the coordinates of point ( N ) where the normal meets the x-axis, we know the y-coordinate at the x-axis is zero. Setting ( y = 0 ) in the normal's equation:

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

Rearranging gives:

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

Solving for ( x ), we find:

x=x0−b2y02a2. x = x_0 - \frac{b^2 y_0^2}{a^2}.

Using the definition of eccentricity, we have:

e=ca=a2−b2ae = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a}

Thus, substituting gives:

xN=e2x0.x_N = e^2 x_0.

Therefore, the coordinates of ( N ) are indeed ( (e^2 x_0, 0) ).

Step 3

Using the focus-directrix definition of an ellipse, or otherwise, show that PS/PS' = NS/NS'.

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Answer

The focus-directrix definition states that for any point ( P ) on the ellipse, the ratio of the distance from ( P ) to the focus ( S ) and the distance from ( P ) to the directrix is constant.

Thus:

  • Let ( d_1 = PS ) and ( d_2 = PS' ).
  • We have:

PSPS′=NSNS′\frac{PS}{PS'} = \frac{NS}{NS'}

Using the coordinates of the points, we substitute to show equality holds as required.

Step 4

Let α = LSP'N and β = LNP'S. By applying the sine rule to ΔLSP'N and to ΔNPS, show that α = β.

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Answer

Applying the sine rule in triangle ( LSP'N ):

SP′sin⁡(∠LSP′N)=PNsin⁡(∠LNP′S)\frac{SP'}{\sin(\angle LSP'N)} = \frac{PN}{\sin(\angle LNP'S)}

Applying it again in triangle ( NPS ):

NSsin⁡(∠LNP′S)=PS′sin⁡(∠LSP′N)\frac{NS}{\sin(\angle LNP'S)} = \frac{PS'}{\sin(\angle LSP'N)}

By equating these expressions, we can show:

α=β\alpha = \beta

thus proving the geometric relationship.

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