Parabola Maths Question Help
The point P(at^2, 2at), t>0 lies on the parabola M with equation y^2=4ax, a>0
a)Show that the equation for the tangent M at P is ty=x+at^2
How do I do this question?
a)Show that the equation for the tangent M at P is ty=x+at^2
How do I do this question?
Original post by SpeakNoEvil
The point P(at^2, 2at), t>0 lies on the parabola M with equation y^2=4ax, a>0
a)Show that the equation for the tangent M at P is ty=x+at^2
How do I do this question?
a)Show that the equation for the tangent M at P is ty=x+at^2
How do I do this question?
Have you done differentiation so you can fin the equation of a tangent to a curve
Yes but I've messed it up and I don't know how
do you want a solution or an explanation?
Explanation.
You different the y^2=4ax equation to then find out the equation that will give you the gradient of a tangent at any point on the curve.
Explanation.
You different the y^2=4ax equation to then find out the equation that will give you the gradient of a tangent at any point on the curve.
Original post by SpeakNoEvil
Yes but I've messed it up and I don't know how
differentiate the equation of the parabola (answer in terms of a)
Evaluate gradient function at the point given
use that gradient and the coordinates of the point given
A bit of algebra and hopefully you will get your answer.
If you did all this post your workings in a photo, so someone can check
Original post by AwesomeSauce#1
do you want a solution or an explanation?
Explanation.
You different the y^2=4ax equation to then find out the equation that will give you the gradient of a tangent at any point on the curve.
Explanation.
You different the y^2=4ax equation to then find out the equation that will give you the gradient of a tangent at any point on the curve.
I'd like a solution so I can cross reference it with my working to see where I've gone wrong
Original post by SpeakNoEvil
I'd like a solution so I can cross reference it with my working to see where I've gone wrong
It is definitely against forum rules for a member to post a full solution.
terribly sorry
Original post by TeeEm
It is definitely against forum rules for a member to post a full solution.
terribly sorry
terribly sorry
O...kaaay can you just tell me what you get when you differentiate y^2=4ax?
First Step is to differentiate y^2=4ax with respect to x.
next is to plug in the x coordinate of the point P to then find the gradient of the tangent at P. from there you should be able to figure it out
next is to plug in the x coordinate of the point P to then find the gradient of the tangent at P. from there you should be able to figure it out
Original post by SpeakNoEvil
O...kaaay can you just tell me what you get when you differentiate y^2=4ax?
if you use implicit differentiation 2y dy/dx = 4a
so dy/dx = (2a)/y
if you use C1 type differentiation
dy/dy = +/- √a x-1/2
Okay, okay I know what I did now
Original post by SpeakNoEvil
Okay, okay I know what I did now
hope it works
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