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If θ is the measure of acute angle between the pair of lines given by `ax^2+2hxy+by^2=0,` then prove that `tantheta=|(2sqrt(h^2-ab))/(a+b)|,a+bne0`

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#### Solution

Let ` m_1 and m_2` be the slopes of the lines represented by the equation

`ax^2+2hxy+by^2=0 .........(1)`

Then their separate equation are

`y=m_1x and y=m_2x`

therefore their combined equation is

`(y - m_1x)(y - m_2x)=0`

`" i.e " m_1m_2x^2-(m_1+m_2)xy+y^2=0 ..................(2)`

Since (1) and (2) represent the same two lines, comparing the coefficients, we get

`(m_1m_2)/a = 1/b = (m_1+m_2)/(2h) `

`therefore m_1+m_2=-(2h)/b and m_1m_2=a/b`

Thus `(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2`

`(m_1 - m_2)^2 = (- (2h)/b)^2 - 4 (a/b)`

`=(m_1 - m_2)^2 = (4(h^2-ab))/b^2`

Let the angle between y = m_{1}x and y =m_{2}x be θ .

`tan theta =|(m_1-m_2)/(1+m_1m_2)|, `

`= |(sqrt (m_1 - m_2)^2)/(1 + m_1m_2)|`

`=|((4sqrt(h^2-ab))/b^2)/(1+a/b)|,`

`therefore tan theta=|(2sqrt(h^2-ab))/(a+b)|,if a + b ne 0`

Hence to proved.

#### RELATED QUESTIONS

Find ‘k' if the sum of slopes of lines represented by equation x^{2}+ kxy - 3y^{2} = 0 is twice their product.

If θ is the acute angle between the lines represented by equation ax^{2} + 2hxy + by^{2} = 0 then prove that `tantheta=|(2sqrt(h^2-ab))/(a+b)|, a+b!=0`

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