Question

the locus of the point represented by x= cos ^2 t,y= 2sint.can anyone pls ans it fast it is urgent n pls don't ans if u don't know.​

Answer 1

Answer:

y^2 = -4 ( x - 1 )

Step-by-step explanation:

x= cos ^2t ; y = 2sin t

we use trigonometric identity I....so

y^2 = (2sin t)^2 = 4 sin^2t.....(1)

now 4x = 4(cos^2t) = 4 cos^2t......(2)

(1) + (2).......

4x + y^2:= 4 ( sin^2t + cos^2t)

4x + y^2:= 4 ×1

4x + y^2:= 4 ,.......(3)

eqn (3) can be written as ....

y^2 = 4 - 4x

y^2 = -4 ( x - 1 ).......(4)

eqn (4) represents a parabola ( y - h)^2 = 4a (x-h) with eccentricity a = -1

therefore the locus ( parabola ) is....

y^2 = -4 ( x - 1 )