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A point moves so that the distance betwe...

A point moves so that the distance between the foot of perpendiculars from it on the lines `a x^2+2hx y+b y^2=0` is a constant `2d` . Show that the equation to its locus is `(x^2+y^2)(h^2-a b)=d^2{(a-b)^2+4h^2}dot`

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From point `p(alpha , beta),` perpendiculars are dropped on the given pair of lines.
According to the question , AB=2d. Clearly OAPB is a cyclic quadrilateral and OP will be the diameter of the circumcircle of this quadrilateral.
`OP =sqrt(alpha^(2)+ beta^(2))`
Let Q be the center of the circle.
therefore , in `DeltaQRB`,
`sintheta=(BR)/(QB)=(2d)/(sqrt(alpha^(2)+beta^(2)))` (1)
Also , angle between the lines is
`tantheta=(2sqrt(h^(2)-b))/(a+b)` (2)
Therefore , from (1) and (2), we get
`(2sqrt(h^(2)-ab))/(a+b)=(2d)/(sqrt(alpha^(2)+beta^(2)-4d^(2)))`
Therefore , the locus of `P(alpha,beta)`is
`(x^(2)+y^(2))(h^(2)-ab)=d^(2){(a-b)^(2)4h^(2)}`
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