Show that the curvex = 2 cos t, y = 3 sin t cos t has two tangents at (0, 0) and find their equations.

This is the case of parametric derivatives in which x and y are expressed as functions of a variable t.

The first derivative implies that:

[tex] \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } [/tex]

So we need to find both derivatives as functions of the variable t, then:

[tex]\frac{dy}{dt} = 3[cos(t)cos(t)+sin(t)(-sin(t))][/tex]
[tex]\frac{dy}{dt} = 3[cos^{2} (t)-sin^{2} (t)][/tex]

[tex]\frac{dx}{dt} = -2sin(t)[/tex]

Thus:

[tex]\frac{dy}{dx} = \frac{-3[cos^{2} (t)-sin^{2} (t)]}{2sin(t)}[/tex]

At (0,0) [tex]x=0[/tex] and [tex]y=0[/tex], so:

[tex] \left \{ {{0=2cos(t)} \atop {0=3sin(t)cos(t)}} \right.[/tex]
∴[tex] \left \{ {{cos(t)=0} \atop {sin(t)cos(t)=0}} \right.[/tex]

There are two values between -π and π which satisfy these equations simultaneously, namely:
[tex]t =[/tex] π/2
[tex]t =[/tex] -π/2

The equation of a straight line given a point and its slope is
[tex]y - y_{0} = m(x- x_{0} )[/tex]

Given that the point is (0,0), then the equation can be written as:
[tex]y = mx [/tex]

So we will find two straight lines:
[tex] \left \{ {{y= m_{1}x } \atop {y= m_{2}x }} \right. [/tex]

For [tex]t = \pi /2[/tex]:

[tex] m_{1}= \frac{dy}{dx} = \frac{-3[cos^{2} ( \pi /2)-sin^{2} ( \pi /2)]}{2sin( \pi /2)} = \frac{3}{2} [/tex]

For [tex]t = -\pi /2[/tex]:

[tex]m_{2}= \frac{dy}{dx} = \frac{-3[cos^{2} ( -\pi /2)-sin^{2} ( -\pi /2)]}{2sin(- \pi /2)} = -\frac{3}{2} [/tex]

Lastly:
[tex]\left \{ {{y= \frac{3}{2} x } \atop {y= -\frac{3}{2} x }} \right. [/tex]