Question

The equation to the locus of a point Pfor
which the distance from P to (-4, 0) is double the
distance from P to x-axis is ...
1) x2 + 3y2 + 8x + 16 = 0
2) x2 + 3y2 - 8x + 16 = 0
3) x2 - 3y2 + 8x - 16 = 0
A) x2 - 3y2 + 8x + 16 = 0
4 x2 - 3y +​

Answer 1

Answer:

Answer

Answer(h+4)2+(k−0)2=(2k)2

Answer(h+4)2+(k−0)2=(2k)2h2−3k2+8h+16=0

Answer(h+4)2+(k−0)2=(2k)2h2−3k2+8h+16=0x2−3y2+8x+16=0

Answer 2

Answer:

Step-by-step explanation:

Let the coordinates of point P ≡(h,k)

By the given condition, we have,

$\sqrt{(h+4)^{2}+(k-0)^{2})}=2k$

$\implies (h+4)^{2}+{k}^{2}=4{k}^{2}$

$\implies {h}^{2}+8h+16+{k}^{2}=4{k}^{2}$

$\implies {h}^{2}-3{k}^{2}+8h+16=0$

Hence the required locus is ${x}^{2}-3{y}^{2}+8x+16=0$