Q104 / 155 · WBJEE 2026 · Mathematics
Let $Z_1, Z_2$ be the roots of the equation $Z^2 + pZ + q = 0$, where the coefficients $p$ and $q$ may be complex numbers and also let $A, B$ represent $Z_1, Z_2$ respectively in the complex plane. If $\angle AOB = \alpha \neq 0$ and $OA = OB$, where $O$ is the origin, then the value of $\dfrac{p^2}{q}\sec^2\dfrac{\alpha}{2}$ will be
- A$\dfrac{1}{4}$
- B$\dfrac{3}{4}$
- Correct answerC$4$
- D$1$
Solution
Let the roots of the equation be $Z_1$ and $Z_2$.
Since $OA = OB$, we have $|Z_1| = |Z_2| = r$.
Given $\angle AOB = \alpha$, the roots can be related as $Z_2 = Z_1 e^{i\alpha}$.
From the quadratic equation $Z^2 + pZ + q = 0$, the sum and product of the roots are:
$Z_1 + Z_2 = -p$
$Z_1 Z_2 = q$
Evaluating the expression $\dfrac{p^2}{q}$:
$\dfrac{p^2}{q} = \dfrac{(-(Z_1 + Z_2))^2}{Z_1 Z_2} = \dfrac{Z_1^2 + Z_2^2 + 2Z_1 Z_2}{Z_1 Z_2} = \dfrac{Z_1}{Z_2} + \dfrac{Z_2}{Z_1} + 2$
Substituting $Z_2 = Z_1 e^{i\alpha}$:
$\dfrac{p^2}{q} = e^{-i\alpha} + e^{i\alpha} + 2$
Using Euler's formula $e^{i\alpha} + e^{-i\alpha} = 2\cos\alpha$:
$\dfrac{p^2}{q} = 2\cos\alpha + 2 = 2(1 + \cos\alpha)$
Using the half-angle identity $1 + \cos\alpha = 2\cos^2\dfrac{\alpha}{2}$:
$\dfrac{p^2}{q} = 4\cos^2\dfrac{\alpha}{2}$
Therefore:
$\dfrac{p^2}{q}\sec^2\dfrac{\alpha}{2} = 4\cos^2\dfrac{\alpha}{2} \sec^2\dfrac{\alpha}{2} = 4$
Answer: $4$
Since $OA = OB$, we have $|Z_1| = |Z_2| = r$.
Given $\angle AOB = \alpha$, the roots can be related as $Z_2 = Z_1 e^{i\alpha}$.
From the quadratic equation $Z^2 + pZ + q = 0$, the sum and product of the roots are:
$Z_1 + Z_2 = -p$
$Z_1 Z_2 = q$
Evaluating the expression $\dfrac{p^2}{q}$:
$\dfrac{p^2}{q} = \dfrac{(-(Z_1 + Z_2))^2}{Z_1 Z_2} = \dfrac{Z_1^2 + Z_2^2 + 2Z_1 Z_2}{Z_1 Z_2} = \dfrac{Z_1}{Z_2} + \dfrac{Z_2}{Z_1} + 2$
Substituting $Z_2 = Z_1 e^{i\alpha}$:
$\dfrac{p^2}{q} = e^{-i\alpha} + e^{i\alpha} + 2$
Using Euler's formula $e^{i\alpha} + e^{-i\alpha} = 2\cos\alpha$:
$\dfrac{p^2}{q} = 2\cos\alpha + 2 = 2(1 + \cos\alpha)$
Using the half-angle identity $1 + \cos\alpha = 2\cos^2\dfrac{\alpha}{2}$:
$\dfrac{p^2}{q} = 4\cos^2\dfrac{\alpha}{2}$
Therefore:
$\dfrac{p^2}{q}\sec^2\dfrac{\alpha}{2} = 4\cos^2\dfrac{\alpha}{2} \sec^2\dfrac{\alpha}{2} = 4$
Answer: $4$
Hence, the correct answer is option: 'C'.