Question

if one of the line given by ax^2+2hxy+by^2=0 is perpendicular to px+ qy=0 then show that ap^2+2hpq+by^2=0​

Answer 1

$\textbf{To prove:}$

$\boxed{a\,p^2+2h\,pq+b\,q^2=0}$

$\text{Let the slopes of the pair of straight lines ax^2+2hxy+by^2=0 }$

$\text{be m_1 and m_2 }$

$\text{Then,}$

$m_1+m_2=\dfrac{-2h}{b}\;\;\text{and}\;\;m_1\,m_2=\dfrac{a}{b}$

$\text{Slope of the line px+qy=0 is \dfrac{-p}{q} }$

$\text{But one of the lines of ax^2+2hxy+by^2=0 is perpendicular to px+qy=0}$

$\implies\,m_1=\dfrac{q}{p}$

$\text{Now,}$

$m_1+m_2=\dfrac{-2h}{b}\;\text{and}\;m_1\,m_2=\dfrac{a}{b}$

$\implies\,\dfrac{q}{p}+m_2=\dfrac{-2h}{b}\;\text{and}\;(\dfrac{q}{p})m_2=\dfrac{a}{b}$

$\implies\,\dfrac{q}{p}+m_2=\dfrac{-2h}{b}\;\text{and}\;m_2=\dfrac{ap}{bq}$

$\implies\,\dfrac{q}{p}+\dfrac{ap}{bq}=\dfrac{-2h}{b}$

$\implies\,\dfrac{bq^2+ap^2}{bpq}=\dfrac{-2h}{b}$

$\implies\,\dfrac{bq^2+ap^2}{pq}=-2h$

$\implies\,bq^2+ap^2=-2h\,pq$

$\implies\boxed{\bf\,ap^2+2h\,pq+bq^2=0}$

Find more:

Find p and q, if the following equation represents a pair of perpendicular lines

2x²+ 4xy – py² + 4x +qy+1=0​

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