Advance Math Help!! (1 Viewer)

viraj30 · Oct 20, 2011

show that for points 'p' and 'q' on the parabola x= 2at, y=at^2 the tangents meet at (a(p+q), apq)

Also, i get confused with letters x, y, p, q, t..are they variables or constants?? god i am lost!!!

help would be much appreciated!

nightweaver066 · Oct 20, 2011

This is 3U mathematics topic under Parametric Equations.

Let P = (2ap, ap^2), Q = (2aq, aq^2) where p and q are parameters which are constants.

Keep in mind that you are looking for a point of intersection between two tangents. This means you are required to find an x and a y value.

$x = 2at, y = at^2$
$x^2 = 4a^2t^2$

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

At P(2ap, ap^2), dy/dx = p
Equation of tangent at P:
$y - ap^2 = p(x - 2ap)$
$y - ap^2 = px - 2ap^2$
$y = px - ap^2...(1)$

Similarly,
Equation of tangent at Q:
$y = qx - aq^2...(2)$

Solving (1) and (2) for intercepts,
$px - ap^2 = qx - aq^2$
$(p -q)x = ap^2 - aq^2$
$(p-q)x = a(p+q)(p-q)$
$x = a(p+q)$
From (1), $y = p(a(p+q)) - ap^2$
$= ap^2 + apq - ap^2$
$= apq$

Therefore the tangents at P and Q meet at (a(p+q), apq)

barbernator · Oct 20, 2011

what do you use to get that form of script nightweaver?

D94 · Oct 20, 2011

barbernator said:
what do you use to get that form of script nightweaver?

Latex. http://community.boredofstudies.org/showthread.php?t=234262 Or you can click the fx button just above where you make your reply/post.

barbernator · Oct 20, 2011

brilliant

good luck in 4u on monday and 3u on wednesday

viraj30 · Oct 20, 2011

nightweaver066 said:
This is 3U mathematics topic under Parametric Equations.

Let P = (2ap, ap^2), Q = (2aq, aq^2) where p and q are parameters which are constants.

Keep in mind that you are looking for a point of intersection between two tangents. This means you are required to find an x and a y value.

$x = 2at, y = at^2$
$x^2 = 4a^2t^2$

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

At P(2ap, ap^2), dy/dx = p
Equation of tangent at P:
$y - ap^2 = p(x - 2ap)$
$y - ap^2 = px - 2ap^2$
$y = px - ap^2...(1)$

Similarly,
Equation of tangent at Q:
$y = qx - aq^2...(2)$

Solving (1) and (2) for intercepts,
$px - ap^2 = qx - aq^2$
$(p -q)x = ap^2 - aq^2$
$(p-q)x = a(p+q)(p-q)$
$x = a(p+q)$
From (1), $y = p(a(p+q)) - ap^2$
$= ap^2 + apq - ap^2$
$= apq$

Therefore the tangents at P and Q meet at (a(p+q), apq)

can you tell what was the point of y= x^2/4a and differentiating it??

nightweaver066 · Oct 20, 2011

viraj30 said:
can you tell what was the point of y= x^2/4a and differentiating it??

In order to find the gradient at that point. Once you have the gradient at that point, you can apply the point-gradient formula in order to obtain the equation of the tangent.

Advance Math Help!! (1 Viewer)

viraj30

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nightweaver066

Well-Known Member

barbernator

Active Member

D94

New Member