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find the equation of the tangent to the curve

y^2 = 4x at the point (16,8)

any help would be much appreciated :smile:
Differentiate equation to find gradient then sub into y-y1=m(x-x1)
(edited 12 years ago)
Reply 2
thanks i can probably go from there
Reply 3
y2=4xy^2=4x

y=±4xy=\pm \sqrt{4x}

Since the tangent touches the graph at (16,8), you only need to worry about the half of the graph above the x-axis, so take:

y=4x=2xy=\sqrt{4x}=2\sqrt{x}

and solve the problem as you would with any similar problem. If you need help actually finding the tangent of a curve at a point I'll help with that :smile:
Reply 4
no it ok, I got it from there, just forgot to differentiate the y :smile:, thanks anyway
Reply 5
although i am stuck on this one question, not to sure how to get it to differentiate
xy=25, at point (5,5) ?
Reply 6
Just rearrange to get it in the form y=something

xy=25xy=25

y=25x=25x1y=\frac{25}{x}=25x^{-1}
(edited 12 years ago)
Reply 7
oh right, thanks for that :smile:
Reply 8
Or you could differentiate implicity using the product rule. And then -rearranging to get dy/dx= ...

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