If the origin is shifted to the point `((a b)/(a-b),0)` without rotation, then the equation `(a-b)(x^2+y^2)-2a b x=0` becomes `(a-b)(x^2+y^2)-(a+b)x y

Question

If the origin is shifted to the point `((a b)/(a-b),0)` without rotation, then the equation `(a-b)(x^2+y^2)-2a b x=0` becomes `(a-b)(x^2+y^2)-(a+b)x y

If the origin is shifted to the point `((a b)/(a-b),0)` without rotation, then the equation `(a-b)(x^2+y^2)-2a b x=0` becomes `(a-b)(x^2+y^2)-(a+b)x y+a b x=a^2` `(a+b)(x^2+y^2)=2a b` `(x^2+y^2)=(a^2+b^2)` `(a-b)^2(x^2+y^2)=a^2b^2`
A. `(a-b)(x^2+y^2)-(a+b)xy+abx=a^2`
B. `(a+b)(x^2+y^2)=2ab`
C. `(x^2+y^2)=(a^2+b^2)`
D. `(a-b)^2(x^2+y^2)=a^2b^2`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
The given equation is
`(a-b)(x^2+y)-2abx=0`
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+(ab)/(a-b)and y=Y+0`
Substituting these in (i), we get
`(a-b)[(X+(ab)/(a-b))^(2) +Y^2]-2ab[X+(ab)/(a-b)]=0`
or `(a-b)[X^(2)+(a^(2)b^(2))/(a-b)^(2)+Y^2+(2abX)/(a-b)]-2abX-(2a^2b^2)/(a-b)=0`
or `(a-b)(X^2+Y^2)=(a^2b^2)/(a-b)`
or `(a-b)^(2)(X^2+Y^2)=a^2b^2`