Given x a , + ,y b = 1 and ax + by = 1 are two variable lines, 'a… - Vidyadip

MCQ

Given $\frac{x}{a}\, + \,\frac{y}{b}= 1$ and $ax + by = 1$ are two variable lines, $'a\ '$ and $'b\ '$ being the parameters connected by the relation $a^2 + b^2 = ab$. The locus of the point of intersection has the equation

✓
$x^2 + y^2 + xy - 1 = 0$
B
$x^2 + y^2 - xy + 1 = 0$
C
$x^2 + y^2 + xy + 1 = 0$
D
$x^2 + y^2 - xy - 1 = 0$

✓

Answer

Correct option: A.

$x^2 + y^2 + xy - 1 = 0$

Let $(h, k)$ be point of intersection then
$\frac{h}{a} + \frac{k}{b} =1$ given ${a^2} + {b^2} = ab$
$\mathop {ah + kb = 1} $
$\Rightarrow \frac{a}{b} + \frac{b}{a} = 1$
multiply ${h^2} + {k^2} + hk(\frac{b}{a} + \frac{a}{b}) = 1$
${h^2} + {k^2} + hk = 1$
${x^2} + {y^2} + xy - 1 = 0$
Note that the locus is not physically viable

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