A) \[\alpha \beta (x+y)=xy(\alpha +\beta )\]
B) \[\alpha \beta (x+y)=2xy(\alpha +\beta )\]
C) \[(\alpha +\beta )(x+y)=2\alpha \beta xy\]
D) None of these
Correct Answer: B
Solution :
The equation of a line passing through the intersection of straight lines \[\frac{x}{\alpha }+\frac{y}{\beta }=1\] and \[\frac{x}{\beta }+\frac{y}{\alpha }=1\] is \[\left( \frac{x}{\alpha }+\frac{y}{\beta }-1 \right)+\lambda \left( \frac{x}{\beta }+\frac{y}{\alpha }-1 \right)=0\] or \[x\,\left( \frac{1}{\alpha }+\frac{\lambda }{\beta } \right)+y\left( \frac{1}{\beta }+\frac{\lambda }{\alpha } \right)-\lambda -1=0\] This meets the axes at \[A\text{ }\left( \frac{\lambda +1}{\frac{1}{\alpha }+\frac{\lambda }{\beta }},0 \right)\] and \[B\text{ }\left( 0,\frac{\lambda +1}{\frac{1}{\beta }+\frac{\lambda }{\alpha }} \right)\]. Let (h, k) be the mid point of AB, then \[h=\frac{1}{2}.\frac{\lambda +1}{\frac{1}{\alpha }+\frac{\lambda }{\beta }},k=\frac{1}{2}.\frac{\lambda +1}{\frac{1}{\beta }+\frac{\lambda }{\alpha }}\] Eliminating \[\lambda \]from these two, we get \[2hk(\alpha +\beta )=\alpha \beta (h+k)\]. \ The locus of \[(h,k)\]is \[2xy(\alpha +\beta )=\alpha \beta (x+y)\].
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