A curve C is defined by the parametric equations: #x=t^2# and #y = t^3-3t#, how do you show that C has two tangents at the point (3,0) and find their equations?

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To show that curve C has two tangents at the point (3,0) and find their equations, we can follow these steps:

1. Calculate the derivatives of x and y with respect to t, denoted by dx/dt and dy/dt.
2. Determine the value of t for which x = 3, since x = t^2.
3. Substitute the value of t obtained in step 2 into the equation for y to find the corresponding y-coordinate.
4. Calculate the derivatives of x and y with respect to t at the point (3,0), denoted by dx/dt and dy/dt.
5. Find the slopes of the tangents at the point (3,0) using the derivatives calculated in step 4.
6. Write the equations of the tangents using the point-slope form, where the point is (3,0) and the slope is determined in step 5.

Let's proceed with the calculations:

1. ( \frac{{dx}}{{dt}} = 2t ) and ( \frac{{dy}}{{dt}} = 3t^2 - 3 ).
2. To find when ( x = 3 ), solve ( t^2 = 3 ). This yields ( t = \pm \sqrt{3} ).
3. Substitute ( t = \sqrt{3} ) and ( t = -\sqrt{3} ) into the equation ( y = t^3 - 3t ) to find the corresponding y-coordinates.
4. Calculate ( \frac{{dx}}{{dt}} ) and ( \frac{{dy}}{{dt}} ) at ( t = \sqrt{3} ) and ( t = -\sqrt{3} ).
5. Evaluate ( \frac{{dy}}{{dx}} ) at ( t = \sqrt{3} ) and ( t = -\sqrt{3} ) to find the slopes of the tangents.
6. Write the equations of the tangents using the point-slope form with the point (3,0) and the slopes calculated in step 5.

After following these steps, you'll have the equations of the two tangents passing through the point (3,0) on the curve C.