Please will someone talk through this?
talk through what?
Adding the questions now :)
oh ok sorry for interrupting
|dw:1425848475996:dw|
i don't get it.
Figure 2 shows the curve with the equation y^2 = 9(x-1) and the line with the equation 3x+y=9
The curve crosses the x-axis at point A, and the line intersects the curve at the points P and Q (a) Find the coordinates of point A (b) Find, using algebra, the coordinates of P and Q
So far I got that, y^2 = 9x-9 so y=3x-3
HINT: to find point A, you know that along the x-axes we have y=0, so solve:\[y^2=0=9(x-1)\]
there is a graph...
The one I drew isn't accurate, just a sketch
@Green101 - use the hint I gave to solve part a)
I got a = (1,0) and I know thanks!!
It is mainly part B I don't understand!!
great! now for part (b) you know that along the line we have:\[3x+y=9\]\[\therefore y=9-3x\]\[\therefore y^2=(9-3x)^2=81-54x+9x^2\]
and we know that along the curve we have:\[y^2=9(x-1)=9x-9\]these two equations must be equal at points P and Q
therefore you need to solve:\[81-54x+9x^2=9x-9\]
you should be able to rearrange this into a standard quadratic equation in \(x\) and then factor it and solve
You are an absolute legend! Thank you so so much! If I post the answer here will you check it??
sure - and thank you very much for the compliment :)
So far I got that \[x = 5, 2 \]
perfect! :)
from \[x^2-7x+10=0\]
So what would I do from here to calculate P and Q?
simplest is to use the equation of the straight line for each x to get the corresponding y value
So, finally I got
\[y=(-3)(5)+9 \] \[\therefore y=6 < 0 \] and therefore Q = (5, -6)
correct co-ordinates for Q but I believe you meant to pu:\[y=-6\lt0\]
*put
\[y=(-3)(2)+9 \]\[\therefore y=3 > 0\] and so p = (2,3)
And yeah I did XD
you have the right answers :)
Yay! Again thank you so much!! Really needed someone to talk through all this!
yw :)
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