Question

3.
If A(2, 0) and B(0, 3) are two points, find
the equation of the locus of point P such
that AP = 2BP.​

Answer 1

Answer:

$-3x ^{2} -3y ^{2}-2x  +8y -32=0$.

Step-by-step explanation:

If two points $A(x1,y1)$ and B $(x2,y2)$ are given, then we applying the distance formula.

Let locus of point $P(α,β)$ ,  

Given, $AP=2BP$  

taking square both sides,  

$AP^{2} =4BP^{2}$

 

now, AP=  

so, $AP² = (α -2)² + (β -0 )²$

 similarly, PB =  

so, $PB² =[tex](α - 0)^{2}  + (β - 3)^{2}$

Then,

$AP^{2} =4BP^{2}(α -2)^{2}  + (β -0 )^{2} =4[ (α - 0)^{2}  + (β - 3)^{2}]$

$-3\alpha ^{2} -3\beta ^{2}-2\alpha +8\beta -32=0$

Now, put $\alpha  = x$ and $\beta  = y$

Thus, $-3x ^{2} -3y ^{2}-2x  +8y -32=0$  is the locus of point P e.g., circle