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Q.Find the equation of the locus of PP, if A=(4,0)A = (4, 0), B=(−4,0)B = (-4, 0) and ∣PA−PB∣=4|PA - PB| = 4.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2025Subjective· 4mImportance★★★★★
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∣PA−PB∣=|PA-PB|= constant describes a hyperbola with foci A,BA,B.

A=(4,0)A=(4,0), B=(−4,0)B=(-4,0), and P=(x,y)P=(x,y) satisfies ∣PA−PB∣=4|PA-PB|=4.

Here the foci are (±4,0)(\pm 4,0) so c=4c=4; and 2a=4⇒a=22a = 4 \Rightarrow a=2. This is exactly the focus-difference definition of a hyperbola with center at the origin (midpoint of A,BA,B) and transverse axis along the x-axis.

b2=c2−a2=16−4=12b^2 = c^2 - a^2 = 16 - 4 = 12

The locus is the hyperbola: …

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