If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by $2x^2 + bxy - 3y^2 = 0$ and the other lines represented by them are perpendicular then $a^2+b^2 =$

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Alternate Solution

Let's find the value of $a^2 + b^2$ when the two quadratic equations share a common line.

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We have two quadratic equations:

$2x^2 + axy + 3y^2 = 0$

$2x^2 + bxy - 3y^2 = 0$

Where:

One line from the first equation coincides with one line from the second equation

The remaining lines are perpendicular to each other

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For a quadratic equation $Ax^2 + Bxy + Cy^2 = 0$, if it represents two lines with slopes $m_1$ and $m_2$, then:

$Am_1m_2 = C$

$(A(m_1+m_2)) = B$

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For the first equation $2x^2 + axy + 3y^2 = 0$:

$2m_1m_2 = 3$

$2(m_1+m_2) = a$

For the second equation $2x^2 + bxy - 3y^2 = 0$:

$2n_1n_2 = -3$

$2(n_1+n_2) = b$

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Let's say $m_1 = n_1$ (one line coincides).

Then:

$m_1m_2 = \dfrac{3}{2}$ and $m_1n_2 = -\dfrac{3}{2}$

This means $m_2 = \dfrac{3}{2m_1}$ and $n_2 = -\dfrac{3}{2m_1}$

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For the other lines to be perpendicular: $m_2 \cdot n_2 = -1$

Substituting: $\dfrac{3}{2m_1} \cdot (-\dfrac{3}{2m_1}) = -1$

$-\dfrac{9}{4m_1^2} = -1$

$\dfrac{9}{4m_1^2} = 1$

$m_1^2 = \dfrac{9}{4}$

$m_1 = \pm \dfrac{3}{2}$

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Calculating $a$ and $b$ using $m_1 = \dfrac{3}{2}$:

$a = -2(m_1+m_2) = -2(\dfrac{3}{2}+\dfrac{3}{2 \cdot \dfrac{3}{2}}) = -2(\dfrac{3}{2}+\dfrac{3}{3}) = -2(\dfrac{3}{2}+1) = -5$

$b = -2(n_1+n_2) = -2(\dfrac{3}{2}-\dfrac{3}{2 \cdot \dfrac{3}{2}}) = -2(\dfrac{3}{2}-1) = -1$

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Therefore: $a^2 + b^2 = (-5)^2 + (-1)^2 = 25 + 1 = 26$