Find locus of points given by `x=a cos^(2)theta,y=b sin^(2)theta`

If x=a cos^(2)theta,y=b sin^(2)theta find (d^2y)/dx^2

x=b sin^(2)theta & y=a cos^(2)theta

Find locus of curve defined by x=a cos theta, y=b sin theta

Find the equation of normal to the curve x = a cos^(3)theta, y=b sin^(3) theta" at point "'theta'.

Find the blanks. (i) If x=a cos^(3)theta,y=b sin^(3)theta then ((x)/(a))^(2/3)+((y)/(b))^(2/3)=ul(P) (ii) if x=a sec theta cos phi,y=b sec theta sin phi and z=c tan theta then (x^(2))/(a^(2))+(y^(2))/(b^(2))-(z^(2))/(c^(2))=ul(Q) (iii) If cos A+cos^(2)A=1 ,then sin^(2)A+sin^(4)A=ul(R)

Find the slope of the normal to the curve x=a sin^(2) theta, y=b cos^(3)theta" at" theta=(pi)/(4) .

The locus of the point whose co-ordinates are x = 3 cos theta + 2 , y = 3 sin theta - 4 , where theta is a parameter , is

x=a cos theta,y=b sin theta then find dy/dx