Another parametrics qu (1 Viewer)

VenomP · Mar 17, 2009

Two points P(2ap, ap^2) and Q(2aq, q^2) lie on the parabola x^2 = 4ay.

i. Show that the equation of the tangent to the parabola at P is y = px - ap^2.

ii. The tangent at P and the line through Q parallel to the y axis intersect at T. Find the coordinawtes of T.

iii. Write down the coordinates of M, the midpoint of PT.

iv. Determine the locus of M when pq = -1.

kurt.physics · Mar 17, 2009

VenomP said:
Two points P(2ap, ap^2) and Q(2aq, q^2) lie on the parabola x^2 = 4ay.

i. Show that the equation of the tangent to the parabola at P is y = px - ap^2.

$Proof$

$x^2 = 4ay$

$\therefore y= \frac{x^2}{4a}$

$\frac{dy}{dx} = \frac{2x}{4a}$

$= \frac{x}{2a}$

$$At$\,\,P(2ap,\,ap^2)$

$\frac{dy}{dx} = \frac{2ap}{2a}$

$= p$

$The equation formula is$

$\,\, y - y_{1} = m(x - x_{1})$

$\therefore y - ap^2 = p(x - 2ap)$

$y - ap^2 = px - 2ap^2$

$y = px - ap^2$

kurt.physics · Mar 17, 2009

VenomP said:
ii. The tangent at P and the line through Q parallel to the y axis intersect at T. Find the coordinawtes of T.

$$The equation of the line through Q parallel to the y axis is$\,\,x = 2aq$

$$We know that the equation of the tangent is$\,\,y=px - ap^2$

$If we solve simultaniously, we can find the co-ordinate of the intersection, ie point T$

$x = 2aq\;\,.\,\,.\,\,.\,\,.\, (1)$

$y = px - ap^2\;\,.\,\,.\,\,.\,\,.\, (2)$

$y = p(2aq) - ap^2\;\;\;\;($By substituting (1) into (2)$)$

$y = 2aqp - ap^2$

$\therefore\; $the coordinates of T are$\;\; (2aq,\,2aqp - ap^2)$

$ie,$

$\fbox{ T(2aq,\,2aqp - ap^2) }$

kurt.physics · Mar 17, 2009

VenomP said:
iii. Write down the coordinates of M, the midpoint of PT.

$P(2ap,\,ap^2)$

$T(2aq,\,2apq - ap^2)$

$M(x,y) = \left( \frac{ x_{1} + x_{2} }{2},\, \frac{ y_{1} + y_{2} }{2} \right)$

$= \left( \frac{2ap +2aq}{2},\, \frac{ ap^2 + 2apq - ap^2 }{2} \right)$

$= \left( \frac{2(ap +aq) }{2},\, \frac{ 2(apq) }{2} \right)$

$= \left(ap +aq,\,apq \right)$

$Therefore the co-ordinates of the Midpoint are,$

$\fbox{M\left(ap +aq\;,\,apq \right)}$

Trebla · Mar 18, 2009

For iv:
$x = a(p + q)\\y=apq\Rightarrowy=-a$ as $pq=-1\\$So, the locus is just the line $ y = -a$

Another parametrics qu (1 Viewer)

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