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  • A variable straight line drawn through the point of intersection of the lines <img alt="\mathrm{\frac{x}{a}+\frac{y}{b}=l \: and \: \frac{x}{b}+\frac{y}{a}=1}" src="https://entrancecorner.oncodecog

A variable straight line drawn through the point of intersection of the lines \mathrm{\frac{x}{a}+\frac{y}{b}=l \: and \: \frac{x}{b}+\frac{y}{a}=1} meets the coordinate axes in \mathrm{A\: and\: B}. The locus of the midpoint of \mathrm{AB} is the curve \mathrm{k x y(a+b)=a b(x+y) where \: k=}

 

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

-2


Answers (1)

best_answer

Any line through the point of intersection of given lines is

\mathrm{ \left(\frac{x}{a}+\frac{y}{b}-1\right)+\lambda\left(\frac{x}{b}+\frac{y}{a}-1\right)=0 }
This meets the \mathrm{ x-axis \: at \: A \equiv\left[\frac{a b(1+\lambda)}{a \lambda+b}, 0\right] }
And meets the \mathrm{y-axis \: at \: B \equiv\left[0, \frac{a b(l+\lambda)}{a+b \lambda}\right]}

Let mid point of \mathrm{A B \: be\: \mathrm{P}(h, k)} then

\mathrm{ \quad h=\frac{a b(1+\lambda)}{2(a \lambda+b)}, k=\frac{a b(1+\lambda)}{2(a+b \lambda)}, \quad \Rightarrow \quad }\mathrm{\frac{1}{k}=\frac{2(a+b \lambda)}{a b(I+\lambda)} }

\mathrm{ \quad \frac{1}{h}=\frac{2(a \lambda+b)}{a b(I+\lambda)} }

\mathrm{\Rightarrow \quad \frac{1}{h}+\frac{1}{k}=\frac{2}{a b(I+\lambda)}[a \lambda+b+a+b \lambda]=\frac{2}{a b(I+\lambda)}[a(\lambda+l)+b(I+\lambda)] }

\mathrm{ =\frac{2(a+b)}{a b} \quad \Rightarrow \quad 2 h k(a+b)=a b(h+k) }

Locus of  \mathrm{P(h, k) \text { is } 2 x y(a+b)=a b(x+y)}

Hence option 2 is correct.



 

Posted by

seema garhwal

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