# Parametric equations

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The parametric equations of the curve C are x=at^2 and y=2at, where a is a postive constant. The points P and Q lie on C and have parameters p and q respectively.

a)Simplifying your answer in each case find

i)The gradient of the tangent to C at the the point P (Which i have done and got 1/p)

ii)The equation of the tangent to C at the point P(Which i have done and got yp-x-ap^2=0)

b)i)Find an expression, in its simplest form, for the gradient of the line PQ.

ii)Explain how you could use the answer of (b)(i) to derive the gradient of the tangent to C at the point P.

Can someone help solve part b i) and b ii)

a)Simplifying your answer in each case find

i)The gradient of the tangent to C at the the point P (Which i have done and got 1/p)

ii)The equation of the tangent to C at the point P(Which i have done and got yp-x-ap^2=0)

b)i)Find an expression, in its simplest form, for the gradient of the line PQ.

ii)Explain how you could use the answer of (b)(i) to derive the gradient of the tangent to C at the point P.

Can someone help solve part b i) and b ii)

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#5

(Original post by

The parametric equations of the curve C are x=at^2 and y=2at, where a is a postive constant. The points P and Q lie on C and have parameters p and q respectively.

a)Simplifying your answer in each case find

i)The gradient of the tangent to C at the the point P (Which i have done and got 1/p)

ii)The equation of the tangent to C at the point P(Which i have done and got yp-x-ap^2=0)

b)i)Find an expression, in its simplest form, for the gradient of the line PQ.

ii)Explain how you could use the answer of (b)(i) to derive the gradient of the tangent to C at the point P.

Can someone help solve part b i) and b ii)

**Ayaz789**)The parametric equations of the curve C are x=at^2 and y=2at, where a is a postive constant. The points P and Q lie on C and have parameters p and q respectively.

a)Simplifying your answer in each case find

i)The gradient of the tangent to C at the the point P (Which i have done and got 1/p)

ii)The equation of the tangent to C at the point P(Which i have done and got yp-x-ap^2=0)

b)i)Find an expression, in its simplest form, for the gradient of the line PQ.

ii)Explain how you could use the answer of (b)(i) to derive the gradient of the tangent to C at the point P.

Can someone help solve part b i) and b ii)

^{2},2ap) and Q(aq

^{2},2aq)

find the gradient of the chord

then let q tend to p, ie set q=p

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(Original post by

Let P(ap

find the gradient of the chord

then let q tend to p, ie set q=p

**TeeEm**)Let P(ap

^{2},2ap) and Qaq^{2},2aq)find the gradient of the chord

then let q tend to p, ie set q=p

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#7

(Original post by

I still dont understand it sorry :/

**Ayaz789**)I still dont understand it sorry :/

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(Original post by

do you know how to find the gradient of 2 points if you have their coordinates?

**TeeEm**)do you know how to find the gradient of 2 points if you have their coordinates?

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#9

(Original post by

Yes you do y2-y1/x2-x1 but we dont have them in integer form? Or am i just acting ******ed

**Ayaz789**)Yes you do y2-y1/x2-x1 but we dont have them in integer form? Or am i just acting ******ed

If the points P and Q lie on the parabola with parametric equations

x = at

^{2 }y = 2at

then

t = p (say) at P and t = q at Q

You can find he gradient of PQ for:

P(ap

^{2},2ap) and Q(aq

^{2},2aq)

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(Original post by

what integer form?

If the points P and Q lie on the parabola with parametric equations

x = at

then

t = p (say) at P and t = q at Q

You can find he gradient of PQ for:

P(ap

**TeeEm**)what integer form?

If the points P and Q lie on the parabola with parametric equations

x = at

^{2 }y = 2atthen

t = p (say) at P and t = q at Q

You can find he gradient of PQ for:

P(ap

^{2},2ap) and Q(aq^{2},2aq)
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#11

(Original post by

What im saying is dont you have to do 2aq-2ap/aq^2-ap^2 ?? How would i do that? And would it not be 1 considering p & q are the same? Im sorry because i dont understand you

**Ayaz789**)What im saying is dont you have to do 2aq-2ap/aq^2-ap^2 ?? How would i do that? And would it not be 1 considering p & q are the same? Im sorry because i dont understand you

first find a simplified expression for the gradient, assuming P and Q are distinct points

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(Original post by

no

first find a simplified expression for the gradient, assuming P and Q are distinct points

**TeeEm**)no

first find a simplified expression for the gradient, assuming P and Q are distinct points

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#13

(Original post by

And how would i do that without y2-y1/x2-x1?

**Ayaz789**)And how would i do that without y2-y1/x2-x1?

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(Original post by

of course with it!

**TeeEm**)of course with it!

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#15

(Original post by

Thats what i said before lol? What values would i use though? You said simplified?

**Ayaz789**)Thats what i said before lol? What values would i use though? You said simplified?

then find the gradient PQ

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#17

(Original post by

Im sorry but i dont understand it:/ that is a past paper question so can you please tell me how to get 4 marks on b i) and bii)

**Ayaz789**)Im sorry but i dont understand it:/ that is a past paper question so can you please tell me how to get 4 marks on b i) and bii)

Since we have defined a general point P (the parametric equations) on the curve, we can sub these values as and or and .

On a side note, it would be best if you could at least attempt the question. TeeEm does not take kindly to people just asking for answers

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(Original post by

Since we have defined a general point P (the parametric equations) on the curve, we can sub these values as and or and .

On a side note, it would be best if you could at least attempt the question. TeeEm does not take kindly to people just asking for answers

**Louisb19**)Since we have defined a general point P (the parametric equations) on the curve, we can sub these values as and or and .

On a side note, it would be best if you could at least attempt the question. TeeEm does not take kindly to people just asking for answers

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**Louisb19**)

Since we have defined a general point P (the parametric equations) on the curve, we can sub these values as and or and .

On a side note, it would be best if you could at least attempt the question. TeeEm does not take kindly to people just asking for answers

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#20

(Original post by

Nope i cant do it:/ give me the starting line?

**Ayaz789**)Nope i cant do it:/ give me the starting line?

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