Let a≠b be two non-zero real numbers. Then the elements of the set A={z∈C:Re(az2+bz)=a and Re(bz2+az)=b} lie on a conic with eccentricity e then

Let z=x + i y Re(az2+bz)=a ⇒a(x2−y2)+bx=a ........(I)Re(bz2+az)=b⇒b(x2−y2)+ax=b........(II)(I) + (II) and (I) − (II) ⇒ x2−y2−x=1 and x2−y2+x=1⊇⇒x2−y2=1

Let a≠b be two non-zero real numbers. Then the elements of the set A={z∈C:Re(az2+bz)=a and Re(bz2+az)=b} lie on a conic with eccentricity e then

Let z=x + i y Re(az2+bz)=a ⇒a(x2−y2)+bx=a ........(I)Re(bz2+az)=b⇒b(x2−y2)+ax=b........(II)(I) + (II) and (I) − (II) ⇒ x2−y2−x=1 and x2−y2+x=1⊇⇒x2−y2=1

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Let $a \neq b$ be two non-zero real numbers. Then the elements of the set $A = {z \in C : Re (a z^{2} + b z) = a a n d Re (b z^{2} + a z) = b}$ lie on a conic with eccentricity e then

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$e^{2} = 1$

$e^{2} = 3$

$e^{2} = 2$

$2 e^{2} = 1$

answer is C.

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Detailed Solution

Let z=x + i y
$\begin{array}{l} Re (a z^{2} + b z) = a \Rightarrow a (x^{2} - y^{2}) + b x = a ........ (I) \\ Re (b z^{2} + a z) = b \Rightarrow b (x^{2} - y^{2}) + a x = b ........ (I I) \\ (I) + (I I) a n d (I) - (I I) \end{array}$
$\begin{array}{l} \Rightarrow x^{2} - y^{2} - x = 1 a n d x^{2} - y^{2} + x = 1 \supseteq \\ \Rightarrow x^{2} - y^{2} = 1 \end{array}$