 Find the combined equation of the images of the pair of lines represented by $a{x^2} + 2hxy + b{y^2} = 0$ in the line mirror y=0. Verified
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Hint: We will take two slopes of line as ${m_1}$ and ${m_2}$ respectively and we have the product of both the slope as $\dfrac{a}{b}$ whereas the summation of both the slope is $\dfrac{{ - 2h}}{b}$ . So, we will find the equation for $y = {m_1}x$ which makes an angle ${\theta _1}$ with y=0 and again its image in the mirror line y=0 makes an angle $- {\theta _1}$ with x-axis. After that we will find the equation for the $y = {m_2}x$ and then we will combine both the equations.

Let $y = {m_1}x$ and $y = {m_2}x$ be two lines represented by $a{x^2} + 2hxy + b{y^2} = 0$ We have the product of both the slope as $\dfrac{a}{b}$ whereas the summation of both the slope as $\dfrac{{ - 2h}}{b}$ respectively.
If we have $y = {m_1}x$ makes an angle ${\theta _1}$ with y=0 (axis) then its image in the line mirror y=0 makes an angle $- {\theta _1}$ with x-axis. So, its equation is given by
$y = \tan ( - {\theta _1})x$ or $y = - \tan ({\theta _1})x$ or $y = - {m_1}x$ ------(i)
So now we will find the equation of the image of $y = {m_2}x$ in y=0
Hence the equation of the image of $y = {m_2}x$ in y=0 is given by $y = - {m_2}x$ ----(ii)
Now we will find the combined equation of the images
So now equation (i) can be written as $y + {m_1}x = 0$ so as equation(ii) can be written as $y + {m_2}x = 0$ respectively
So let’s find the final solution
$(y + {m_1}x)(y + {m_2}x) = 0$
Now let’s multiply its individually with each term to get the final answer
$y(y + {m_2}x) + {m_1}x(y + {m_2}x) = 0$
Now we have ${y^2} + {m_2}xy + {m_1}xy + {m_1}{m_2}{x^2} = 0$
Now we will take out xy from second $b{y^2} - 2hxy + a{x^2} = 0$ and third term respectively,
${y^2} + xy({m_2} + {m_1}) + ({m_1}{m_2}){x^2} = 0$
Now here we have the product of both the slope as $\dfrac{a}{b}$ whereas the summation of both the slope as $\dfrac{{ - 2h}}{b}$ .
So in place of ${m_1}{m_2}$ we will write $\dfrac{a}{b}$ and ${m_1} + {m_2}$ we will write $\dfrac{{ - 2h}}{b}$
So after replacing these values we got that
${y^2} + xy\left( {\dfrac{{ - 2h}}{b}} \right) + \dfrac{a}{b}{x^2} = 0$
So now we have
So, the correct answer is “ ${y^2} + xy\left( {\dfrac{{ - 2h}}{b}} \right) + \dfrac{a}{b}{x^2} = 0$ ”.