Question

If the slope of one of the lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be the square of the other, then $\dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}$ is equal to
A. \[{{a}^{2}}b+a{{b}^{2}}-6abh+8{{h}^{3}}=0\]
B. \[{{a}^{2}}b+a{{b}^{2}}+6abh+8{{h}^{3}}=0\]
C. \[{{a}^{2}}b+a{{b}^{2}}-3abh+8{{h}^{3}}=0\]
D. \[{{a}^{2}}b+a{{b}^{2}}-6abh-8{{h}^{3}}=0\]

Answer 1

Hint: For a straight line represented by $y=mx+c$ the slope of the straight line is $m$. Through the value of slope we can also get the equation of the straight line.

Formula used: $a{{x}^{2}}+2hxy+b{{y}^{2}}$

Complete step by step solution: The given equation is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
Now let the slope of one of the pairs of straight lines is $m$. Given that the slope of the other straight line is square of the one line. Thus the slope of the other line will be ${{m}^{2}}$ .
Now the equation of the straight lines in terms of slope can be given as-
$(y-mx)(y-{{m}^{2}}x)=0$
Simplifying the equation we get-
$
   {{y}^{2}}-myx-{{m}^{2}}yx+{{m}^{3}}{{x}^{2}}=0 \\
  {{y}^{2}}-xy\left( m+{{m}^{2}} \right)+{{m}^{3}}{{x}^{2}}=0 \\
$
Comparing it with the given equation of the straight line that is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$we get-
$a={{m}^{3}},b=1,h=-\dfrac{\left( m+{{m}^{2}} \right)}{2}$
Putting the value of $a,b,h$ in $\dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}$ we get-
$
   \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2({{m}^{3}}+1)}{-\left( m+{{m}^{2}} \right)}+\dfrac{8{{\left( m+{{m}^{2}} \right)}^{2}}}{4{{m}^{3}}} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2({{m}^{3}}+1)}{-\left( m+{{m}^{2}} \right)}+\dfrac{8{{m}^{2}}{{\left( 1+m \right)}^{2}}}{4{{m}^{3}}} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2({{m}^{3}}+1)}{-\left( m+{{m}^{2}} \right)}+\dfrac{2{{\left( 1+m \right)}^{2}}}{m} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2(m+1)\left( 1-m+{{m}^{2}} \right)}{-m\left( 1+m \right)}+\dfrac{2{{\left( 1+m \right)}^{2}}}{m} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2\left( 1-m+{{m}^{2}} \right)}{-m}+\dfrac{2{{\left( 1+m \right)}^{2}}}{m} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2\left( -1+m-{{m}^{2}}+1+{{m}^{2}}+2m \right)}{m} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=\dfrac{2\times \left( 3m \right)}{m} \\
  \dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}=6 \\
  {{a}^{2}}b+a{{b}^{2}}-6abh+8{{h}^{3}}=0 \\
$
Thus we can write that if the slope of one of the lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be the square of the other, then $\dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}$ is equal to \[{{a}^{2}}b+a{{b}^{2}}-6abh+8{{h}^{3}}=0\].

Thus, Option (A) is correct.

Note: The other method to solve this equation is to determine the sum of the two slopes which is equal to $\dfrac{-2h}{b}$ and multiplication of the two slopes which is equal to $\dfrac{a}{b}$ . From the general form of straight line and comparing $\dfrac{a+b}{h}+\dfrac{8{{h}^{2}}}{ab}$ we can get the values of $a,b,h$ .