Parametrics (1 Viewer)

Skeptyks · Sep 21, 2011

If a question says 'derive the equation of the tangent to the curve $x^{2}=4ay$
at the point (2ap, ap^2), does that mean to go through the whole process or can i just chuck the equation straight in?

Alkanes · Sep 21, 2011

Usually its 2 marks for this question. So process would do. If it was 1 mark just write down the equation.

Skeptyks · Sep 21, 2011

Thanks also, for the question "Find the equation of the chord XY on the parabola x^2 = 8y where X = (4t, 2t^2) and Y = (4r, 2r^2)".
Now, the equation of the chord is $y-\frac{p+q}{2}x+apq=0$ but I have no idea how to put that information into it so instead, I should find the gradient and then use the point gradient formula blablabla?
EDIT: Solved dw but still need an answer from below.

SpiralFlex · Sep 21, 2011

Skeptyks said:
If a question says 'derive the equation of the tangent to the curve $x^{2}=4ay$
at the point (2ap, ap^2), does that mean to go through the whole process or can i just chuck the equation straight in?

You need to derive it unless it says "DO NOT prove this result."

Skeptyks · Sep 21, 2011

Ahh k thanks, got it ^^. For that question just above, what do I do?

EDIT2: Since it says find the equation, I have to make it into cartesian form correct? (I don't have the answers on me ><)

SpiralFlex · Sep 21, 2011

The answer should be in terms of $t$ and $r$ . Use your method you have just suggested. You should get:

$y-\frac{1}{2} (r+t)x +2tr=0$

Skeptyks · Sep 21, 2011

Aww man, i got x after that bracket, is that wrong?

SpiralFlex · Sep 21, 2011

Skeptyks said:
Aww man, i got x after that bracket, is that wrong?

Spiral is at fault here sorry. I am very tired today.

Skeptyks · Sep 21, 2011

ahh it is all good! Thanks

Skeptyks · Sep 21, 2011

One more question, find the equation of the chord of contact of tangents drawn from the external point (3,-1) to the parabola $x^{2}=8y$ .
I know the equation is $xx_{1}=2a(y+y_{1})$ but I have no clue how to derive it.

SpiralFlex · Sep 21, 2011

I drew a pretty diagram for you.

Let's find the gradient:

$=\frac{ap^2-aq^2}{2ap-2apq}$

$=\frac{a(p+q)(p-q)}{2a(p-q)}$

$=\frac{1}{2}(p+q)$

Now,

$y-aq^2=\frac{1}{2}(p+q)(x-2aq)$

$y-aq^2=\frac{1}{2}(p+q)x - aq(p+q)$

$y=\frac{1}{2}(p+q)x -apq$

We can prove that the tangents intersect at this point. If you want I can prove by differentiating it first then finding equations for tangents at P and Q but I am sure you know how to do that, so for now accept it.

$(a(p+q), apq)$

Let's define the external intersection point as $(x_{1}, y_{1})$ . We can see that they are the exact same points.

$x_{1}=a(p+q)$

$y_{1}=apq$

Now,

$(p+q)=\frac{x_{1}}{a}$

So we can substitute these values into our original equation.

$y=\frac{1}{2}(\frac{x_{1}}{a})x -y_{1}$

$2ay= x x_{1}- 2a y_{1}$

$2a(y+y_{1} )=x x_{1}$

Skeptyks · Sep 22, 2011

SpiralFlex said:
I drew a pretty diagram for you.

Let's find the gradient:

$=\frac{ap^2-aq^2}{2ap-2apq}$

$=\frac{a(p+q)(p-q)}{2a(p-q)}$

$=\frac{1}{2}(p+q)$

Now,

$y-aq^2=\frac{1}{2}(p+q)(x-2aq)$

$y-aq^2=\frac{1}{2}(p+q)x - aq(p+q)$

$y=\frac{1}{2}(p+q)x -apq$

We can prove that the tangents intersect at this point. If you want I can prove by differentiating it first then finding equations for tangents at P and Q but I am sure you know how to do that, so for now accept it.

$(a(p+q), apq)$

Let's define the external intersection point as $(x_{1}, y_{1})$ . We can see that they are the exact same points.

$x_{1}=a(p+q)$

$y_{1}=apq$

Now,

$(p+q)=\frac{x_{1}}{a}$

So we can substitute these values into our original equation.

$y=\frac{1}{2}(\frac{x_{1}}{a})x -y_{1}$

$2ay= x x_{1}- 2a y_{1}$

$2a(y+y_{1} )=x x_{1}$

Thankyou! Took me a bit to figure out how to derive it with the given co-ordinate but I think I got it now

Thanks for spending the time + effort.

SpiralFlex · Sep 22, 2011

Skeptyks said:
Thankyou! Took me a bit to figure out how to derive it with the given co-ordinate but I think I got it now Thanks for spending the time + effort.

Anything for a delicious, I mean delightful human.

Skeptyks · Sep 22, 2011

SpiralFlex said:
Anything for a delicious, I mean delightful human.

Hehe, I got 3x-4y+4=0 in the end. Does that sound about right o-o?

SpiralFlex · Sep 22, 2011

Skeptyks said:
Hehe, I got 3x-4y+4=0 in the end. Does that sound about right o-o?

Correct human. Spiral will commence shut down mode now.

Skeptyks · Sep 22, 2011

SpiralFlex said:
Correct human. Spiral will commence shut down mode now.

Okay haha, goodnight.

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