Prove that the single equation of two lines which pass through the original and perpendicular to the pair of lines represented by the equation ax² + 2hxy +by²=0 is bx² -2hxy +ay²= 0.

Solution:
Given,

Single equation of straight lines is ax²+2hxy+by²=0
[Dividing by b]
$or, \frac{a}{b}x^2 + \frac{2h}{b}xy + \frac{b}{b}y^2 = 0$
$or, \frac{a}{b}x^2 + \frac{2h}{b}xy + y^2 = 0$ - (i)

Let the two separate equations of straight lines represented by the given single equation be:
($y = m_1x$) -(1) and ($y = m_2x$) - (2)

Combining two lines, we get,
$(m_1x -y)(m_2x -y) = 0$
$or, m_1m_2x^2 - m_1xy -m_2xy +y^2 = 0$
$or, m_1m_2x^2 -(m_1+m_2)xy +y^2 = 0$ - (ii)

Comparing equation (ii) with equation (i), we get,
$m_1m_2 = \frac{a}{b}$ and $m_1+m_2= - \frac{2h}{b}$
 
Two line perpendicular to the liens in equation (1) and (2) are: ($m_1y + x= 0$) and ($m_2y +x = 0$)

Combining these two equations of liens, we get,
$or, (m_1y +x)(m_2y +x) = 0$
$or, m_1m_2y^2 + (m_1+m_2)xy + x^2 = 0$
[Put value of $m_1m_2$ and $m1+m_2$ from above]
$or, \frac{a}{b}y^2 + - \frac{2h}{b}xy + x^2 = 0$
[Multiplying by b]
$or, \frac{a}{b}by^2 - \frac{2h}{b}bxy + bx^2 = 0$
$or, ay^2 - 2hxy +bx^2 = 0$
$or, bx^2 - 2hxy +ay^2 = 0$ is the required equation.

Therefore, the required single equation of two straight lines is bx² -2hxy +ay² = 0. #proved.

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