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 $z=z_a{\cos }^{4}t+2z_b{\cos }^{2}t{\sin }^{2}t+z_c{\sin }^{4}t(t\in R)$
A line bisecting AB and parallel to AC intersects the given curve at

• A. Two distinct points
• B. Two co-incident points
• C. Only one point
• D. No point

Answer 1

Answer: (C)

Only one point

$z_a=i\Rightarrow A(0,1)$
$z_b=\frac{1}{2}+2i\Rightarrow B(\frac{1}{2},2)$
$z_c=1+4i\Rightarrow C(1,4)$
Let D be the midpoint of AB
$D=[\frac{0+\frac{1}{2}}{2},\frac{1+2}{2}]$
$D[\frac{1}{4},\frac{3}{2}]$
Line bisecting AB at D is parallel to AC
$\therefore$ Slope of AC $=\frac{4-1}{1-0}=3$
Equation of line bisecting AB is $y-\frac{3}{2}=3(x-\frac{1}{4})$

$\Rightarrow \frac{2y-3}{2}=\frac{12x-3}{4}$
$\Rightarrow 4y-6=12x-3$
$\Rightarrow 12x-4y+3=0$
$\Rightarrow y=\frac{12x+3}{4}$
Equation of curve is $y={(x+1)}^2$
$\frac{12x+3}{4}=x^2+2x+1$
$4x^2-4x+1=0$
${(2x-1)}^2=0$
$x=\frac{1}{2},y=\frac{9}{4}$
$(\frac{1}{2},\frac{9}{4})\Rightarrow$ given line and curve intersect only at one point