Question

Let $A\left(z_a\right),B\left(z_b\right),C\left(z_c\right)$ are three non-collinear points where $z_a=i,z_b=\frac{1}{2}+2i,z_c=1+4i$ and a curve is $z=z_a{\cos }^{4}t+2z_b{\cos }^{2}t{\sin }^{2}t+z_c{\sin }^{4}t(t\in R)$
Equation of curve in cartesian form

• A. $y=x^2+x+1$
• B. $y=(x+1{)}^2$
• C. $y=(x-1{)}^2$
• D. $y=x^2+x-1$

Answer 1

The correct option is B $y=(x+1{)}^2$

$z_a=i,z_b=\frac{1}{2}+2i,z_c=1+4i$
Let $z=x+iy$
$z=z_a{\cos }^{4}t+2z_b{\cos }^{2}t{\sin }^{2}t+z_c{\sin }^{4}t$
$z=i{\cos }^{4}t+2[\frac{1}{2}+2i]{\cos }^{2}4{\sin }^{2}4+(1+4i){\sin }^{4}t$
$z=i{\cos }^{4}t+{\cos }^{2}t{\sin }^{2}t+4i{\cos }^{2}t{\sin }^{2}t+{\sin }^{4}t+4i{\sin }^{4}t$
$z=i{\cos }^{4}t+{\sin }^{2}t({\cos }^{2}t+{\sin }^{2}t)+4i{\sin }^{2}t({\cos }^{2}t+{\sin }^{2}t)$
$z=i{\cos }^{4}t+{\sin }^{2}t+4i{\sin }^{2}t$
$z={\sin }^{2}t+i({\cos }^{4}t+4{\sin }^{2}t)$
$x+iy={\sin }^{2}t+i({\cos }^{4}t+4{\sin }^{2}t)$
Compare real and imaginary parts
$x={\sin }^{2}t⟶1$
$y={\cos }^{4}t+4{\sin }^{2}t$
$y={\left({\cos }^{2}t\right)}^2+4{\sin }^{2}t$
$y={(1-{\sin }^{2}t)}^2+4{\sin }^{2}t$
$y=1+{\sin }^{4}t-2{\sin }^{2}t+4{\sin }^{2}t$
$y={\left({\sin }^{2}t\right)}^2+2{\sin }^{2}t+{\left(1\right)}^2$
$y={[{\sin }^{2}t+1]}^2$
From equation 1
$y={(x+1)}^2$