HELP~ I suck at MX1 - crap student here. (1 Viewer)

RivalryofTroll · Dec 23, 2012

Diagram (unavailable) shows point P(t,t^2) and Q (1-t, (1-t)^2) on parabola y = x^2. R is intersection of tangents to parabola at P & Q.

(I've done the previous parts to the question I'm asking so I'll skip those)

Given that tangents intersect at R(1/2, t-t^2) (already proved it previously),

Describe the locus of R as t-varies, stating any restriction on the y-coordinate.

So is it x=1/2, a vertical straight line? What are the restrictions?

RealiseNothing · Dec 23, 2012

SpiralFlex · Dec 23, 2012

RealiseNothing said:
$x=\frac{1}{2}$

$y \leq \frac{1}{4}$

Saw that edit. Not quite. Don't encourage Rivalry to rote learn pls.

RealiseNothing · Dec 23, 2012

SpiralFlex said:
Saw that edit. Not quite. Don't encourage Rivalry to rote learn pls.

my head hurts way too much to care.

SpiralFlex · Dec 23, 2012

$First of all, we need to verify that our intuition is correct. Firstly, by testing some tangents we can successful say that it would be impossible to find a locus point inside the parabola.$

$Secondly, how do we derive an answer of$\; x=\frac{1}{2}? \;$Well, we a point of intersection is$\; R(\frac{1}{2}, t-t^2) \forall t \in \mathbb{R}. \;$We found that to be the intersection for all real$\; t\; $values, so the locus is the fixed line at$\; x=\frac{1}{2}. \;$Now, we need to derive at our restriction.$

$Consider the point \;$R(\frac{1}{2}, t-t^2).$ We said t can be any value correct? Well, the maximum value it can be found.$

$y=t-t^2$

$y'=1-2t=0$

$t=\frac{1}{2}$

$y=\frac{1}{4}$

$So the maximum value on a parabola (this one) is\;$ \frac{1}{4}. \;$But does that mean the locus is$\; x=\frac{1}{2}\; $for$ \; y \leq \frac{1}{4}?$We notice that,$\; (\frac{1}{2}, \frac{1}{4})\;$lies on the parabola, so it can't be an intersection point. The locus R is therefore,$

$x=\frac{1}{2}, y < \frac{1}{4}$

Also don't beat yourself up like that...

RivalryofTroll · Dec 23, 2012

SpiralFlex said:
$First of all, we need to verify that our intuition is correct. Firstly, by testing some tangents we can successful say that it would be impossible to find a locus point inside the parabola.$

$Secondly, how do we derive an answer of$\; x=\frac{1}{2}? \;$Well, we a point of intersection is$\; R(\frac{1}{2}, t-t^2) \forall t \in \mathbb{R}. \;$We found that to be the intersection for all real$\; t\; $values, so the locus is the fixed line at$\; x=\frac{1}{2}. \;$Now, we need to derive at our restriction.$

$Consider the point \;$R(\frac{1}{2}, t-t^2).$ We said t can be any value correct? Well, the maximum value it can be found.$

$y=t-t^2$

$y'=1-2t=0$

$t=\frac{1}{2}$

$y=\frac{1}{4}$

$So the maximum value on a parabola (this one) is\;$ \frac{1}{4}. \;$But does that mean the locus is$\; x=\frac{1}{2}\; $for$ \; y \leq \frac{1}{4}?$We notice that,$\; (\frac{1}{2}, \frac{1}{4})\;$lies on the parabola, so it can't be an intersection point. The locus R is therefore,$

$x=\frac{1}{2}, y < \frac{1}{4}$

Also don't beat yourself up like that...

Thanks Spiral.

SpiralFlex · Dec 23, 2012

For a more optimistic outlook change your thread title to "developing student".

HELP~ I suck at MX1 - crap student here. (1 Viewer)

RivalryofTroll

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RealiseNothing

what is that?It is Cowpea

SpiralFlex

Well-Known Member

RealiseNothing

what is that?It is Cowpea

SpiralFlex

Well-Known Member

RivalryofTroll

Sleep Deprived Entity

SpiralFlex

Well-Known Member

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