(i) x = at^{2} , y = 2at represent the equation of parabola y^{2} = 4ax
(ii) x = acosθ , y = bsinθ represent the equation of ellipse 

+ 

= 1 
(iii) x = asecθ , y = btanθ represent the equation of hyperbola 

+ 

= 1 
Solution
(i) x = at^{2} , y = 2at represent the equation of parabola y^{2} = 4ax
The parametric equations are:
x = at^{2} ..........(i)
y = 2at ..........(ii)
From (ii) y = 2at
By taking square on both sides
y^{2} = (2at)^{2}
y^{2} = 4a^{2}t^{2}
y^{2} = 4a(at^{2})
So according to (i) x = at^{2}
y^{2} = 4ax which is the equation of parabola
(ii) x = acosθ , y = bsinθ represent the equation of ellipse 

+ 

= 1 
The parametric equations are:
x = acosθ ..........(i)
y = bsinθ ..........(ii)
From (i) x = acosθ
By taking square on both sides
( 

)^{2} = cos^{2}θ ..........(a) 
From (ii) y = bsinθ
By taking square on both sides
( 

)^{2} = sin^{2}θ ..........(b) 
Now by adding equation (a) and (b)
( 

)^{2} 
+ 
( 

)^{2} 
= cos^{2}θ + sin^{2}θ 
As cos^{2}θ + sin^{2}θ = 1

+ 

= 1 which is the equation of ellipse 
(iii) x = asecθ , y = btanθ represent the equation of hyperbola 

+ 

= 1 
The parametric equations are:
x = asecθ ..........(i)
y = btanθ ..........(ii)
From (i) x = asecθ
By taking square on both sides
( 

)^{2} = sec^{2}θ ..........(a) 
From (ii) y = btanθ
By taking square on both sides
( 

)^{2} = tan^{2}θ ..........(b) 
Now by subtracting equation (a) and (b)
( 

)^{2} 
 
( 

)^{2} 
= sec^{2}θ  tan^{2}θ 
As sec^{2}θ  tan^{2}θ = 1

+ 

= 1 which is the equation of ellipse 