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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.

**:**

__Complete step-by-step answer__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

Hence, this is the answer.

**So, the correct answer is “ $ {y^2} + xy\left( {\dfrac{{ - 2h}}{b}} \right) + \dfrac{a}{b}{x^2} = 0 $ ”.**

**Note**: We should always remember the product and summation of the slope and apply it in the equations to get the simplified value. As the question is asked for y=0 line as the mirror uses the fact associated with the x-axis.

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