Use implicit differentiation to find the points of tangency

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SUMMARY

This discussion focuses on using implicit differentiation to find points of tangency for the hyperbola defined by the equation 9x² - y² = 36. The user successfully derived the slope of the tangent line as y' = 9x/y. To find the points of tangency, it is essential to establish the equation of the tangent line at a point (x₀, y₀) and evaluate its intersection with the y-axis. The final expression for the y-coordinate at the y-axis is given by y = (-9x₀² + y₀²) / y₀.

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  • Knowledge of slope-intercept form of a line
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Students studying calculus, particularly those focusing on implicit differentiation and conic sections, as well as educators looking for examples of tangent lines to hyperbolas.

ziddy83
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Please Help!

Ok, I am having problem with an Implicit differentiation problem...
Two tangent lines to the hyperbola [tex]9x^2 - y^2 =36[/tex] intersect at the y-axis.

Use implicit differentiation to find the points of tangency. Ok so i implicitly differentiated this function and i came up with y'= 9x/y...now...Im not sure on how to find the points of tangency. Do i have to set the two equations equal to each other and solve for y and x separately? Please help, and help will be appreciated! thanks
 
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You really don't need to worry about "two" tangent lines. Choose any point on the ellipse, find the tangent line and see where it intersects the y_axis.

Let (x0, y0) be the point of tangency. Then the slope of the tangent line is, as you found, 9x0/y0 and the equation of the tangent line is [itex]y= \frac{9x_0}{y_0}(x- x_0)+ y_0[/itex]. At the y-axis, x= 0 so
[itex]y= \frac{9x_0}{y_0}(-x_0)+ y_0= \frac{-9x_0^2+y_0^2}{y_0}[/itex].
 
Thank you for your response! I was just confused on what to pick the points as. Thanks again
 

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