Solve Hyperbola Homework with Tangent Equation | Graph Included

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The discussion focuses on solving a hyperbola homework problem involving the tangent equation. The tangent line is expressed as y = -x/t^2 + x(t)/t^2 + y(t). Participants clarify that to determine if a tangent can pass through the point (4,4), one must differentiate the hyperbola implicitly to find dy/dx and express it in terms of a single variable. The solution involves setting up two equations based on the tangent line's form and solving for the unknowns, x0 and c.

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http://img184.imageshack.us/img184/6506/hmmyh4.jpg

The question ebfore had me find the equation of the tangent for any parametizaton values of (x(t),y(t)).

Which is y = -x/t^2 + x(t)/t^2 + y(t)

I'm pretty sure the case that a tangent to the hyperbola can't pass through the point is because the point lies inside the curve.

So basically all i have to do is show that for x = 4 in the original equation has a y value less than 4, and for y = 4 in the equation has a x value less than 4? By the way I sketched the graph.Did I do this right?
 
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I don't understand your answer for the parametrisation of the hyperbola, the expression gives me y = y(t), which doesn't say much, to say the least.

I don't think you have to parametrise the curve anyway. You first have to find dy/dx by differentiating implicitly and then suppose there is a tangent line to the curve which passes through (4,4) in the form y= mx + c. Let m=dy/dx. Just make sure you express dy/dx in terms of one variable, either y or x only.

Then you have 2 equations:

4 = 4m + c

y_{0} = mx_{0} + c

where y_{0} and x_{0} is the point on the hyperbola for which a tangent line to it passes through (4,4). As above, you should be able to express y_{0} in terms of x_{0} alone. Then you now have 2 equations and 2 unknowns, x0 and c. Try solving for them and see what happens.

I'm not sure what you're trying to achieve with your quasi-graphical method, being able to sketch the graph helps, but it's entirely possible to do this analytically as above.
 
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