If \[\dfrac{{{x^2}}}{a} + \dfrac{{{y^2}}}{b} + \dfrac{{2xy}}{h} = 0\] represents a pair of straight lines and slope if one line is twice the other, then \[ab:{h^2}\] is
1) \[9:8\]
2) \[8:9\]
3) \[1:2\]
4) \[2:1\]

Hint: After analyzing the pair of straight lines, we can conclude that straight lines pass through the origin, i.e., they are of the form \[y = mx\] . No particular formula is required to solve this question, but an approach to compare two equations having similar parameters is applied.

Complete step-by-step answer:
Now, we are given the pair of straight lines as \[\dfrac{{{x^2}}}{a} + \dfrac{{{y^2}}}{b} + \dfrac{{2xy}}{h} = 0 - - - - (i)\] ,
Let the slope of one of the lines be m,
Then the slope of the other line is 2m (as given in the question),
So, one line is \[y = mx\] , and the other line is \[y = 2mx\] .
Now, to obtain the pair of straight lines using the given lines; we can multiply them as shown below,
 \[ \Rightarrow (y - mx)(y - 2mx) = 0\] ,
Now, after opening the brackets, we get,
 \[ \Rightarrow {y^2} - mxy - 2mxy + 2{m^2}{x^2} = 0\]
Simplifying the equation we get,
 \[ \Rightarrow {y^2} - 3mxy + 2{m^2}{x^2} = 0 - - - - (ii)\]
Now, Let’s compare equation \[(i)\] and \[(ii)\] ,
The coefficient of \[{y^2}\] , \[xy\] and \[{x^2}\] are equated, and we obtain,
 \[a = \dfrac{1}{{2{m^2}}}\] ,
 \[h = - (\dfrac{2}{{3m}})\] ,
 \[b = 1\]
Now, we are required to find \[ab:{h^2}\] , therefore, by substituting their values from the above equations we get,
 \[ \Rightarrow ab:{h^2} = \dfrac{1}{{2{m^2}}} \times 1:{( - (\dfrac{2}{{3m}}))^2}\]
Now, solving the R.H.S
 \[ \Rightarrow ab:{h^2} = \dfrac{1}{{2{m^2}}} \times 1 \times {( - (\dfrac{{3m}}{2}))^2}\]
And finally, we obtain the value of the required ratio as shown below
 \[ \Rightarrow ab:{h^2} = \dfrac{9}{8}\]
Thus, option(1) is the correct answer.
So, the correct answer is “Option 1”.

Note: This question requires one to analyze and inspect the pair of straight lines. Do not commit calculation mistakes, and be sure of the final answer. There is no particular formula for this question but an approach to develop and solve the question.