ax² -2hxy -by² = 0 | Pair Of Straight Lines | SciPiPupil

Question: Find the equation of straight lines passing through the origin and perpendicular to the given line: $ax² -2hxy -by² = 0$.

Solution:

Given,

Equation of a single line is $ax² -2hxy -by² = 0$ - (i)

Let $m_1$ and $m_2$ represent the slopes of the two separate straight lines.

Then the equation of the two straight lines passing through the origin is represented by: 

Line 1: $y = m_1x$

$or, y - m_1x = 0$

And

Line 2: $y=m_2x$

$or, y - m_2x = 0$

Combining the equation of line 1 and line 2, we get,

$or, (y -m_1x)(y - m_2x) = 0$

$or, y (y -m_2x) - m_1x(y -m_2x) = 0$

$or, y^2 - ym_2x - ym_1x + m_1m_2x^2= 0$

$or, y^2 - (m_1+m_2)xy + m_1m_2 x^2 = 0$

$or, m_1m_2x^2 - (m_1+m_2)xy +y^2 - (ii)

From equation (i), we have,

$ax² -2hxy -by² = 0$

[Dividing by b]

$or, \frac{ax^2}{b} - \frac{2h}{b}xy - \frac{by^2}{b} = 0$

$or, \frac{a}{b}x^2 - \frac{2h}{b}xy -y^2 = 0$

$or, - \frac{a}{b}x^2 + \frac{2h}{b}xy + y^2 = 0$ - (iii)

Comparing equation (ii) with equation (iii), we get,

$m_1m_2 = - \frac{a}{b}, (m_1+m_2) = -\frac{2h}{b}$

Equation of lines perpendicular to the lines given by$y - m_1x = 0$ and $y - m_2x = 0$ are $m_1y+x=0$ and $m_2y+x = 0$, respectively.

Combining the new equation of lines, we get,

$or, (m_1y +x)(m_2y +x) = 0$

$or, m_1y(m_2y +x) + x(m_2y +x) = 0$

$or, m_1m_2y^2 + m_1xy + m_2xy +x^2 = 0$

$or, x^2 + (m_1+m_2)xy + m_1m_2y^2 = 0$

[Put value of $m_1m_2$ and $m_1+m_2$ from above]

$or, x^2 - \frac{2h}{b}xy - \frac{a){b}y^2 = 0$

[Multiplying by b]

$or, bx^2 - \frac{2h}{b}bxy-  \frac{a}{b}by^2 = 0$

$or, bx^2 - 2hxy - ay^2 = 0$ is the required equation.

Therefore,the required equation of the straight line is bx² -2hxy -ay² = 0.

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