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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}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%

A variable straight line drawn through the point of intersection of the lines \mathrm{\frac{x}{a}+\frac{y}{b}=1} and \mathrm{\frac{x}{b}+\frac{y}{a}=1}  meets the coordinate axes in 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 \mathrm{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 x-axis at A \mathrm{\equiv\left[\frac{a b(l+\lambda)}{a \lambda+b}, 0\right]}

And meets the \mathrm{ y- axis ~at~ B \equiv\left[0, \frac{a b(1+\lambda)}{a+b \lambda}\right]}

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

\begin{aligned} &\mathrm{ \quad h=\frac{a b(I+\lambda)}{2(a \lambda+b)}, k=\frac{a b(l+\lambda)}{2(a+b \lambda)} \quad \Rightarrow \quad \frac{l}{h}=\frac{2(a \lambda+b)}{a b(I+\lambda)}, }\\ &\mathrm{ \frac{l}{k}=\frac{2(a+b \lambda)}{a b(I+\lambda)} }\\ &\mathrm{ \Rightarrow \quad \frac{l}{h}+\frac{l}{k}=\frac{2}{a b(l+\lambda)}[a \lambda+b+a+b \lambda] \quad \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) }\\ &\mathrm{ \text { Locus of } P(h, k) \text { is } \quad 2 x y(a+b)=a b(x+y)}\\ \end{aligned}

 

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manish painkra

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