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bored of sc
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A tangent to a parabola is a straight line that touches a parabola only once.shaon0 said:
kaz1
et tu
To find a tangent of a parabola differnetiate the equation of the parabola, sub in the x value of the point into the derivative which gives you the gradient of the tangent, then use the gradient in the point gradient formula where y-y1 = m(x-x1)
kaz1
et tu
lolzshaon0 said:
I didn't see the question.
conics2008
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c) A(0,3) and B(-2,1), Given these points. Find the locus of the point P(x,y) so that APB is a right angle.
its a circle.
with eqution (x+1)^2 + (y-2)^2=2
its a circle.
with eqution (x+1)^2 + (y-2)^2=2
np.shaon0 said:
you could do it by finding the gradients of lines AP and BP and beucase its a right angle, they should multiply to give you -1 or using the distance formula, i think they all give the right answer.
3unitz
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find the gradient of line AP:
mAP = (y - 3) / (x - 0)
find the gradient of line BP:
mBP = (y - 1) / (x + 2)
for angle APB to be a right angle both lines have to be perpendicular to each other. this implies:
mAP x mBP = - 1
(y - 3)(y - 1) / (x - 0)(x + 2) = - 1
y^2 - 4y + 3 = - (x^2 + 2x)
y^2 - 4y + 4 - 1 = - (x^2 + 2x + 1 - 1)
(y - 2)^2 - 1 = - [(x + 1)^2 - 1]
(y - 2)^2 + (x + 1)^2 = 2
locus is a circle radius sqrt 2, centered at (-1, 2), excluding the points A and B
the circle geometry method is a lot quicker as you can see from the messy math. always draw a neat sketch as sometimes there's quicker ways to get the equation from the geometry.
mAP = (y - 3) / (x - 0)
find the gradient of line BP:
mBP = (y - 1) / (x + 2)
for angle APB to be a right angle both lines have to be perpendicular to each other. this implies:
mAP x mBP = - 1
(y - 3)(y - 1) / (x - 0)(x + 2) = - 1
y^2 - 4y + 3 = - (x^2 + 2x)
y^2 - 4y + 4 - 1 = - (x^2 + 2x + 1 - 1)
(y - 2)^2 - 1 = - [(x + 1)^2 - 1]
(y - 2)^2 + (x + 1)^2 = 2
locus is a circle radius sqrt 2, centered at (-1, 2), excluding the points A and B
the circle geometry method is a lot quicker as you can see from the messy math. always draw a neat sketch as sometimes there's quicker ways to get the equation from the geometry.
Thanks. is that like calculus?3unitz said:a tangent is a line which "just touches" a parabola (or other curve) with a special type of gradient. the way you find the gradient of a tangent is by bringing 2 points closer and closer together:
View attachment 16750
if you calculate the gradient as the green line gets "really close" to the point, you get the gradient of the tangent at the point
Thanks for the links.3unitz said:that is calculus, yep
there are 2 main branches of calculus: differentiation and integration, both of which are closely related (fundamental theorem of calculus)
finding gradients of tangents comes under differentiation. integration encompasses finding areas and volumes (among a great deal of other things of course)
sites for introduction to differentiation:
http://www.maths.mq.edu.au/numeracy/web_mums/module3/Worksheet38/module3.pdf
http://www.ma.hw.ac.uk/~des/F11EP1/block4.pdf
I'll post my answers for the first part in
http://www.maths.mq.edu.au/numeracy/...38/module3.pdf
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