First slide

Cartesian plane

Question

If the origin is shifted to the point $(\frac{a b}{a - b}, 0)$ without rotation, then the equation $(a - b) (x^{2} + y^{2}) - 2 abx = 0$ becomes

Moderate

A

$(a - b) (x^{2} + y^{2}) - (a + b) xy + abx = a^{2}$
B

$(a - b) (x^{2} + y^{2}) = 2 ab$
C

$(x^{2} + y^{2}) = (a^{2} + b^{2})$
D

${(a - b)}^{2} + (x^{2} + y^{2}) = a^{2} b^{2}$

Solution

The given equation is
$(a - b) (x^{2} + y^{2}) - 2 abx = 0$ ……(1)
The origin is shifted to (ab/(a-b),0). Any point (x.y) on the curve (1) must be replaced with a new point (X,Y) with reference to the new axes, such that
$x = X + \frac{ab}{a - b}$ and $y = Y + 0$
Substituting these in (i), we get
$(a - b) [{(X + \frac{ab}{a - b})}^{2} + Y^{2}] - 2 ab [X + \frac{ab}{a - b}] = 0$
or $(a - b) [{(X^{2} + \frac{a^{2} b^{2}}{{(a - b)}^{2}})}^{2} + Y^{2} + \frac{2 abX}{a - b}] - 2 abX - \frac{2 a^{2} b^{2}}{a - b} = 0$
or $(a - b) (X^{2} + Y^{2}) = \frac{a^{2} b^{2}}{a - b}$
or $(a - b) (X^{2} + Y^{2}) = a^{2} b^{2}$

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