# Tangent line to the curve

## Homework Statement

This is a very basic problem, though it did confuse me a little:

Find the tangent equations to the curve $$x=3t^2+1 \ , \ y = 2t^3+2$$ which intercepts the point (4,3).

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## The Attempt at a Solution

I took $$\frac{dy}{dx} = t = \frac{y-y_0}{x-x_0} = \frac{y-3}{x-4} \rightarrow y=t(x-4)+3$$

What to do next? I don't think the equation works for all t.

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Mark44
Mentor

## Homework Statement

This is a very basic problem, though it did confuse me a little:

Find the tangent equations to the curve $$x=3t^2+1 \ , \ y = 2t^3+2$$ which intercepts the point (4,3).
Are you sure the above is right? If x = 4, then t = +/-1, but when t = 1, y = 4 and when t = -1, y = 0.

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## The Attempt at a Solution

I took $$\frac{dy}{dx} = t = \frac{y-y_0}{x-x_0} = \frac{y-3}{x-4} \rightarrow y=t(x-4)+3$$

What to do next? I don't think the equation works for all t.

Yes, I'm sure, those are the functions.

I believe the problem asks for a tangent of the graphic which will intercept the point (4,3) in R^2, which is not necessarily in the graphic of the function.

Mark44
Mentor
OK, I misunderstood.

So let's say we're talking about the point on the curve whose coordinates are (x0, y0), that correspond to t = t0.

Can you use the parametric equations to write x0 and y0 in terms of t0. Then use the point (4, 3) and calculate the slope of the line segment between (x0, y0) and (4, 3), which you know is equal to t0.