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

x^{2} + y^{2} + x y - 1 = 0

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2 x^{2} + y^{2} - x y - 1 = 0

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2 x^{2} + y^{2} + x y - 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - x y - 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} + y^{2} + x y - 1 = 0$
Solution:-

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Similar questions

(1, 1)

x^{2} + x y - 3 x - y + 2 = 0

x^{2} - y^{2} - 2 x + 2 y = 0

x y - x - y + 1 = 0

x y - y^{2} - x + y = 0

b x t - a y t = a b

b x + a y = a b t

t

List I	List II
$A)$ The locus of the point $(a sec θ, b tan θ)$	$1)$ $x^{2} - y^{2} = a^{2}$
$B)$ The locus of the point $(c t, \frac{c}{t})$	$2)$ $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$C)$ The locus of the point $(a sec θ, a tan θ)$	$3)$ $x y = c^{2}$
$D)$ The locus of the point $(a cos θ, a sin θ)$	$4) x^{2} + y^{2} - a^{2} = 0$

p (θ), D (θ + \frac{π}{2})

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

2 x^{2} - x y - y^{2} + x + 2 y - 1 = 0

x + y + 1 = 0

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