A) 9 : 8
B) 8 : 9
C) 1 : 2
D) 2 : 1
Correct Answer: A
Solution :
Let \[{{m}_{1}},{{m}_{2}}\] be the slopes \[\therefore \]\[{{m}_{1}}+{{m}_{2}}=-\frac{2b}{h}\] and \[{{m}_{1}}{{m}_{2}}=\frac{b}{a}\] Again\[{{m}_{2}}\] =\[2{{m}_{1}}\] \[\therefore \]\[3{{m}_{1}}=-\frac{2b}{h}\] and \[2m_{1}^{2}=\frac{b}{a}\] \[\therefore \]\[\frac{9m_{1}^{2}}{2m_{1}^{2}}=\frac{4{{b}^{2}}}{{{h}^{2}}}\times \frac{a}{b}\Rightarrow ab:{{h}^{2}}=9:8\].
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