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The equations with tangents at (0,0) are [tex]y = \frac{4}{7} x[/tex] and
[tex]y = -\frac{4}{7} x[/tex].
In this question,
The curves are x = 7 cos(t), y = 4 sin(t) cos(t)
Two tangents at (0, 0)
In this case, the parametric derivative of x and y are expressed in terms of t.
The first derivative dy/dx can be expressed as
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
Now, dy/dt is obtained by differentiate y with respect to t,
[tex]\frac{dy}{dt}= 4[cos(t)(cos(t))+sin(t)(-sin(t))][/tex]
⇒ [tex]\frac{dy}{dt}= 4[cos^{2} (t)-sin^{2} (t)][/tex]
Now, dx/dt is obtained by differentiate x with respect to t,
[tex]\frac{dx}{dt} =7(-sin(t))[/tex]
⇒ [tex]\frac{dx}{dt} =-7sin(t)[/tex]
Thus, [tex]\frac{dy}{dx}=\frac{4[cos^{2}(t)-sin^{2}(t ) ]}{-7sin(t)}[/tex]
At (0,0) x = 0 and y = 0, Then
0 = 7 cos(t)
0 = 4 sin(t) cos(t)
and
cos(t) = 0
sin(t) cos(t) = 0
There are two values between -π and π which satisfy these equations simultaneously are
t = π/2
t = -π/2
The equation of a straight line given a point and its slope is
y-y₀ = m(x-x₀)
The two tangents lies at (0,0), so the equation becomes
y = mx
Then the two straight lines will be
y = m₁x and
y = m₂x
For t = π/2,
[tex]m_1=\frac{dy}{dx}=\frac{4[cos^{2}(\frac{\pi }{2} )-sin^{2}(\frac{\pi }{2} ) ]}{-7sin(\frac{\pi }{2} )}[/tex]
⇒ [tex]m_1=-\frac{4[0-1]}{7(1)}[/tex]
⇒ [tex]m_1=\frac{4}{7}[/tex]
For t = -π/2,
[tex]m_2=\frac{dy}{dx}=\frac{4[cos^{2}(-\frac{\pi }{2} )-sin^{2}(-\frac{\pi }{2} ) ]}{-7sin(-\frac{\pi }{2} )}[/tex]
⇒ [tex]m_2=-\frac{4[0-1]}{-7(1)}[/tex]
⇒ [tex]m_2=-\frac{4}{7}[/tex]
Thus the equations with tangents at (0,0) are [tex]y = \frac{4}{7} x[/tex] and
[tex]y = -\frac{4}{7} x[/tex].
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