Question

If one of the lines represented by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be $y=mx$, then another straight line is-
A. \[b{{m}^{2}}+2hm+a=0\]
B. \[b{{m}^{2}}+2hm-a=0\]
C. \[a{{m}^{2}}+2hm+b=0\]
D. \[a{{m}^{2}}-2hm+b=0\]

Answer 1

Hint: Here we have to substitute the value of $y$ in the equation of the pair of straight lines to get the equation of the other line as the equation of the pair of straight lines is a product of the equation of the separate lines.

Formula Used:2. $a{{x}^{2}}+2hxy+b{{y}^{2}} =0$

Complete step by step solution: The given equation of the pair of straight line is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
The equation of one of the line is $y=mx$.
To determine the equation of the another straight line we have to put the value of $y$ in the given equation of the pair of straight lines.
Putting the value of $y$ in the given equation of pair of straight lines we get-

 $ a{{x}^{2}}+2hxy+b{{y}^{2}}=0 $
 $ a{{x}^{2}}+2hx\left( mx \right)+b{{\left( mx \right)}^{2}}=0 $
 $ a{{x}^{2}}+2hm{{x}^{2}}+b{{m}^{2}}{{x}^{2}}=0 $
 $ {{x}^{2}}\left( a+2hm+b{{m}^{2}} \right)=0 $
 $ x=0 $
 $\left( a+2hm+b{{m}^{2}} \right)=0$

Thus the equation of the another line is $\left( a+2hm+b{{m}^{2}} \right)=0$.
So, we can write that if one of the lines represented by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be $y=mx$, then another straight line is \[b{{m}^{2}}+2hm+a=0\].

Option ‘A’ is correct

Note: A pair of straight lines can be real or imaginary types. It can be also coincident or distinct. The nature of a straight line can be determined from the ${{h}^{2}}-ab$ value. If the value of ${{h}^{2}}-ab$ is zero the straight lines are real and coincident and if the value of ${{h}^{2}}-ab$ is greater than zero then the straight lines are real and distinct.