# If θ is the measure of acute angle between the pair of lines given by ax2+2hxy+by2=0, then prove that tanθ=∣(2(h2−ab))/(a+b)∣,a+b≠0 - Mathematics and Statistics

Sum

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

#### 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 = m1x and y =m2x 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.

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2012-2013 (March)

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