2009 Hills Grammar 4u trial, Q4 b) help plz? (1 Viewer)

[darkfenrir]

I forgot what to do...

 

[darkfenrir]

Someone help me please?

 

[asianese]

i) Use y=c^2/x and differentiate. Sub in the line equation with T and you'll get it.

ii) Replace t with p and q in the tangent equation. Solve simultaneously for X and Y.

iii) use X= whatever it is, Y= whatever it is, and manipulate it so that you can incorporate p^2+q^2=2 into it. Simplify and get the answer.

 

[nightweaver066]

i) Differentiate x & y values of T w.r.t "t" and find dy/dx. Then use point-gradient formula to get equation of the tangent.

ii) Replace t with p for equation of tangent at P, and replace t with q for equation of tangent at Q. Solve these two equations simultaneously for R.

iii) You find that R is .

Given that





Multiplying x and y coordinates of R together,

 

[Sy123]

Its just a basic locus question really, I would even classify it as a Harder 3U question.

For part 4bi), what does it say?

'Show that the tangent to the hyperbola at any point T is ..... '

So its asking us to find the tangent to the parabola at point T(ct, c/t), therefore we must find the gradient of the curve at that point (dy/dx)




Therefore the gradient of the curve at the point T, where x=ct is:




This is the gradient , we already have the point

Using point gradient formula:



rearrange and collect like terms:



Part ii)

Now lets use the previous question's identity:

instead of T lets put in P instead and put Q instead:

Equation of line at point P:



Equation of line at point Q



Now in the question it is given that R is the intersection of these 2 tangents, so lets solve them simultaneously
This is easy enough, we arrive at



Sub this into the equation again to get something for x now:



therefore:



Part iii)

Here they give us a relationship between p and q, this may be useful for later so lets write it down:



We are given a show that:

So lets construct this using the co-ordinates of R:





By expanding the denominator and using teh identity given:

Simplify mroe:

 

[darkfenrir]

i) Differentiate x & y values of T w.r.t "t" and find dy/dx. Then use point-gradient formula to get equation of the tangent.

ii) Replace t with p for equation of tangent at P, and replace t with q for equation of tangent at Q. Solve these two equations simultaneously for R.

iii) You find that R is .

Given that





Multiplying x and y coordinates of R together,

thankyou kind sir I got up to multiplying x and y and simplifying, didnt realise about the +2c^2-2c^2 thing haha

 

[darkfenrir]

Its just a basic locus question really, I would even classify it as a Harder 3U question.

For part 4bi), what does it say?

'Show that the tangent to the hyperbola at any point T is ..... '

So its asking us to find the tangent to the parabola at point T(ct, c/t), therefore we must find the gradient of the curve at that point (dy/dx)




Therefore the gradient of the curve at the point T, where x=ct is:




This is the gradient , we already have the point

Using point gradient formula:



rearrange and collect like terms:



Part ii)

Now lets use the previous question's identity:

instead of T lets put in P instead and put Q instead:

Equation of line at point P:



Equation of line at point Q



Now in the question it is given that R is the intersection of these 2 tangents, so lets solve them simultaneously
This is easy enough, we arrive at



Sub this into the equation again to get something for x now:



therefore:



Part iii)

Here they give us a relationship between p and q, this may be useful for later so lets write it down:



We are given a show that:

So lets construct this using the co-ordinates of R:





By expanding the denominator and using teh identity given:

Simplify mroe:

Thanks for your time sy Your and nightweavers explanations are both quite cool, yours is a bit more unorthodox though but it works haha