Question

Find the value of h if the slopes of the lines represented by 6x^2+2hxy+y^2 = 0 are in the ratio 1:2

Answer 1

Answer:

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Answer 2

The value of $h=\dfrac{-3\sqrt{3}}{2}$ .

Explanation:

The equation of a pair of lines passing through origin : $ax^2+2hxy+by^2$

Here , the lines represented by : $y-m_1x=0$ and $y-m_2x=0$.

Then, $m_1+m_2=\dfrac{-2h}{b}$    (1) and $m_1m_2=\dfrac{a}{b}$   (2)

Given equation : $6x^2+2hxy+y^2 = 0$ , here a= 6 and b =1

Also, $\dfrac{m_1}{m_2}=\dfrac{1}{2}\Rightarrow\ m_2=2m_1$

Now , from (2) ,

$m_1m_2=\dfrac{6}{1}\\\\\Rightarrow\ m_1(2m_1)=6\\\\\Rightarrow\ m_1^2=3\\\\\RIghtarrow\ m_1=\sqrt{3}$

Then , $m_2= 2(\sqrt{3})$

Substituent corresponding values in (1) , we get

$\sqrt{3}+2\sqrt{3}=\dfrac{-2h}{1}\\\\\Rightarrow\ -2h=3\sqrt{3}\\\\\Rightarrow\ h=\dfrac{-3\sqrt{3}}{2}$

Hence, the value of $h=\dfrac{-3\sqrt{3}}{2}$ .

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Find the slope of line x/3+y/2=1

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