Really hard parametrics question- need help !! (1 Viewer)

[kooltrainer]

okay, heres the question:

16) the points P and Q on thr parabola x=2at, y=at^2 have parameters p and q respectively.
a) Find the midpoint M of the chord PQ, the point T on the parabola where its tangent is parallet to PQ, and the point I where the tangents at P and Q intersect.
b) Show that M,T and I lie in a vertical line, with T the midpoint of MI
c) Show that the tangent at T bisects the tangents PI and QI
d) What is the ratio of the areas of triangle PQI to triangle PQT?

i've done a) and b) and i need help with c) and d) ! ..
heres a diagram that i drew :]

 

[kony]

they're both quite straight forward really, everything's set out for you.

for c), use the similarity argument. Let's call the place where the tangent at T intersects IP and IQ "H" and "K" respectively.

quite clearly, triangles IHK and IPQ are similar (angle PIQ is common, PQ//HK). since MT = TI, PH = HI and QK = KI.

d) PQ is a common base. Since the height is doubled (MI = 2TI), the area is also doubled, using the area = 1/2.base.height.

 

[Triple777ER]

The equation of the Tangents at P and Q respectively are
y = px - ap^2
y = qx - aq^2

Find the equation of the tangent at T, and then solve it simultaneously with the equations of the tangents at P and Q separately, and then you will find the co-ordinates. Then you can conclude that the tangent at T bisects the tangents PI and QI

Im still doing (d). If you need the answer to (c) clearly, i will try and scan one for you, but have a go at it.

 

[kony]

by the way, fellow fortian, how did you draw that diagram?

i'm also fort st, graduated last year. you'll see me at speech day if you don't skip it.

 

[kooltrainer]

Find the equation of the tangent at T, and then solve it simultaneously with the equations of the tangents at P and Q separately, and then you will find the co-ordinates. Then you can conclude that the tangent at T bisects the tangents PI and QI

.

i know thats gona take 10 years to do.. even finding tangent at T will give u a messy equation.. there must be a straight forward way
but if u've done it , i don't mind looking

 

[kooltrainer]

they're both quite straight forward really, everything's set out for you.

for c), use the similarity argument. Let's call the place where the tangent at T intersects IP and IQ "H" and "K" respectively.

quite clearly, triangles IHK and IPQ are similar (angle PIQ is common, PQ//HK). since MT = TI, PH = HI and QK = KI.

d) PQ is a common base. Since the height is doubled (MI = 2TI), the area is also doubled, using the area = 1/2.base.height.

, ur looking for areas of triangle PQI and triangle PQT. not PQI or HKI .. triangle PQI and triangle PQT are not similar?

.. MTI is not the height of any triangle ...

 

[kony]

yes, it's not the height of a triangle there, but it's pretty obvious what the missing step is, isn't it?

join a perpendicular from I to PQ. That is the height. (i had a diagonal height, which worked because of similarity)

i'm sorry it's not that clear, but i'm trying to be concise in the answer, and not feed you the answer, so you can think about it as well.

this is the straight forward "hence"-way for c) and d) [using similarity and construction]


if you're still unsure, i'll do a handwritten solution.