I cant seem to work out this Parametric problem can anyone help its for tomorrow

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TheAssassin96

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1)A curve has parametric equations x = t2 , y = 4t. ............edited
Find the equation of the normal to this curve at (t2 , 4t). .....edited
Find the coordinates of the points where this normal cuts the coordinate axes.
Hence find in terms of t, the area of the triangle enclosed by the normal and the axes.
 
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You imply that you have made some attempts, and failed. Please show us at least one attempt, including what went wrong. In order to help, we need to know what you know, and where you need help.

In particular, I expect to see you apply some sort of formula or procedure to find the normal; I would expect to see at least that, so we can observe whether you did it all wrong or made some silly mistake. Or maybe you got that part right, but forgot how to find intercepts.

I'm assuming you wrote "t 2" meaning t2, which you can write as t^2.
 
Okay. Now, what were you taught about normal lines? What method are you attempting to follow? (There are several ways you might do it, from using a formula involving vectors down to finding the perpendicular slope and writing the point-slope equation of the line.)
 
You've already done the differentiation part; that's fine.

That tells you the slope of the curve at a given point. What is the slope of the normal line at that point? This is no longer calculus! Give it a try.
 
TheAssassin96 said:
i dont know how .....

can you just pls do it for me....
Click to expand...
Can you tell us "how the slope of the tangent-line" is related to the derivative?

Then

What is the relationship between the slopes of two lines that are normal to each other?
 
No, I will not spoon-feed you. If you can differentiate, then you have learned about the slopes of lines; and if you were given this assignment, then you have been taught what a normal line is. Be a big boy and try. It's the only way to learn.

Have you forgotten that the slope of a line perpendicular to a line with slope m is -1/m? Have you forgotten how to write an equation of a line given a point and a slope? If so, look it up.
 
look this is what i got so far i dont know how i would continue
95099516_523023181726598_8442504195977773056_n.jpg
 
now i got this far..is it good ?
2y=-tx+4t is the normal line
whilst (0,2t)and(4,0) is where the normal line cuts the x and y axis respectively

94977232_285086289178158_3970535340589973504_n.jpg
 
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TheAssassin96 said:
now i got this far..is it good ?
2y=-tx+4t is the normal line
whilst (0,2t)and(4,0) is where the normal line cuts the x and y axis respectively
Click to expand...
Close ...

Writing your line as y = (-t/2)x + 2t, we see that the slope is -t/2, which is what we need; but when x = t^2, the point it is supposed to pass through, we get y = -t^3/2 + 2t^3 = 3/2 t^3, not 4t as it should be.

Looking at your work, I see that the first line is the equation of the tangent, not the normal. Then at the fourth line you just replaced the slope with the perpendicular slope, without correcting anything else -- you kept the same y-intercept as for the tangent line.

So repeat your work, but use the correct slope from the start, and you will probably have it.
 
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