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Tangents

If I’ve found the equation of the tangent to a curve at a point (2,-10) and got y=8x-26. How would I find the coordinates of another point that is parallel to that tangent?
all you need is the gradient, m=8, and change the y-intercept, c, to anything you want (e.g 4). then try a few x-values or simply put the y intercept as a coordinate (0, c)
(edited 5 years ago)
Any line that has the same gradient to that line (a gradient of 8) is parallel to that line. For example the line 8x-20 is parallel to that line. From the info you’ve given I think you could choose any value for the y-intercept and that would be on a line that is parallel to y=8x-26 as long as the gradient of the line is 8.
Reply 3
Avatar for tracey0
tracey0
OP
8
So would the coordinate (0,-26) be parallel?
Original post by tracey0
So would the coordinate (0,-26) be parallel?

That coordinate would actually be on the line you mentioned, y=8x-26, because if you substitute x=0 into the equation you get y=-26. To answer your question, it's not exactly on a parallel line because it's on the line itself.
Reply 5
Avatar for tracey0
tracey0
OP
8
Ohhh okay, I understand now, sorry. So would (4,6) be okay?
(0,26) wouldn’t work (for the reasons stated above) but (0,25) or any other y value would.
With regards to (4,6) it would depend on the equation you have given to the parallel line
Reply 7
Avatar for tracey0
tracey0
OP
8
So if I sub in x=5 for example
y=8(5)-26
y=14

Would (5,14) be parallel?
(edited 5 years ago)
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Original post by tracey0
So if I sub in x=5 for example
y=8(5)-26
y=14

Would (5,14) be parallel?

you're still subbing x=5 into the same line. You have to change the y-intercept of the line you're subbing an x value into, or you'll be finding a point on the line itself and not on a parallel line.
To change the y-intercept, just come up with any random number other than -26.
Example: -24
So then the equation of the line would be: y=8x-24
And a point on this line would be, subbing in x=5 like you did: y=8(5)-24
y=16
So the point (5,16) would be parallel and not on the same line.
Original post by tracey0
If I’ve found the equation of the tangent to a curve at a point (2,-10) and got y=8x-26. How would I find the coordinates of another point that is parallel to that tangent?

It doesn't make sense to ask if a point is parallel to a line (the responses to you have effectively been answering "is your point *on* a line parallel (but not equal to) to a line", and they are all, frankly, missing the point).

The question asks you to find a tangent (line) that is parallel to the line you've found.

Two lines of the form y = mx + c are parallel if they have the same gradient. And the tangent to a curve f(x) has gradient f'(x).

So, to sum it up, you are looking for (the other) point on the curve with the same gradient as the gradient at P.

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