Given: The equation of curve is x = sin3t, y = cos2t.
Here we have to find the equation of the tangent the curve x = sin3t, y = cos2t at t = π/4
When t = π/4, then x = sin (3π/4) = sin [π - (π/4)] = sin π/4
=1∕√2 Similarly, when t = π/4, then y = cos [2 ⋅ (π/4)] = cos π/2 = 0
So, the point of contact is
(,0) As we know that slope of the tangent at any point say
(x1,y1) is given by:
m=[](x1,y1) Now by differentiating x = sin3t, y = cos2t with respect to t we get,
⇒ dx/dt = 3 cos 3t and dy/dt = - 2 sin 2t
As we know that
⇒==g′(t)f′(t) and
==f′(t)g′(t) ⇒==− ⇒[]t==− == As we know that slope normal at any point say
(x1,y1) is given by:
− ⇒ Slope of normal is:
−3∕2√2 As we know that equation of normal at any point say
(x1,y1) is given by:
y−y1 =−[](x1,y1).(x−x1) Here,
x1=1∕√2,y1=0 and
[]t==− ⇒y−0=.(x−) ⇒3√2.x+4y−3=0 Hence, equation of the required tangent is
3√2x+4y−3=0