UPSC NDA Math Model Paper 8

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 (

1

√2

,0)
As we know that slope of the tangent at any point say (x1,y1) is given by: m=[

dy

dx

](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 ⇒

dy

dx

=

dy dt

dx dt

=g′(t)f′(t) and

dx

dy

=

dx dt

dy dt

=f′(t)g′(t)
⇒

dy

dx

=

dy dt

dx dt

=−

2‌sin‌2‌t

3‌cos‌3‌t

⇒[

dy

dx

]t=

π

4

=−

2‌sin( π 2 )

3‌cos( 3π 4 )

=

2‌sin( π 2 )

3‌cos( π 4 )

=

2√2

3

As we know that slope normal at any point say (x1,y1) is given by: −

1

[ dy dx ](x1,y1)

⇒ 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 =−[

dx

dy

](x1,y1).(x−x1)
Here, x1=1∕√2,y1=0 and [

dx

dy

]t=

π

4

=−

3

2√2

⇒y−0=

−3

2√2

.(x−

1

√2

)
⇒3√2.x+4y−3=0
Hence, equation of the required tangent is 3√2x+4y−3=0