Parametric Question for old New Senior Mathematics 3U (1 Viewer)

McRafi9 · Jan 23, 2017

Ex 24(c) Q17. pg. 79 New Senior Mathematics 3UNIT Fitzpatrick

17. Through the vertex A of a parabola, chords AP and AQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola in a fixed point K.

Thanks

dan964 · Jan 24, 2017

Missing some details in some of the steps, but oh well. (finally proofchecked)

Firstly to prove for a specific case for parabola where
$$ Vertex is $ (0,0). $ Focus is $ (0,a)$

$P $ is $ (2ap, ap^2) $ and $ Q $ is $ (2aq, aq^2)$
$$ Gradient of $ AP $ is $ \frac{p}{2}$
$$ Gradient of $ AQ $ is $ \frac{q}{2}$
$\Rightarrow \frac{pq}{4}=-1 $ (using $ AP \bot AQ $)$$
$\Rightarrow pq=-4$
$$ Gradient of $ PQ $ is $ \frac{p+q}{2}$ (if you don't know how to show this ask)
$$ Equation of line $ PQ $ is $ 2y-2ap^2=(p+q)(x-2ap)$
$\Rightarrow 2y=(p+q)x-2apq$

$$ At axis $ x=0$
$\Rightarrow y=-2apq$

$\Rightarrow y=8a $ which is a fixed point $ \box$

Parametric Question for old New Senior Mathematics 3U (1 Viewer)

McRafi9

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dan964

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