Solution: Let P (h, k) represent any point on the locus And let the coordinates of A and B be given by (0, 0) and (h, 0). Where, PA = 3PB Upon squaring both the sides we get, h2 + k2 = 9k2 h2 = 8k2...
Find the locus of a point which is equidistant from (1, 3) and x-axis.
Solution: Let P (h, k) be any point on the locus and let A (1, 3) and B (h, 0). Where, PA = PB
Find the locus of a point such that the sum of its distances from (0, 2) and (0, -2) is 6.
Solution: Let P (h, k) represent any point on the locus And let the coordinates of points A and B be (0, 2) and (0, -2). Where, PA – PB = 6
A point moves as so that the difference of its distances from (ae, 0) and (-ae, 0) is 2a, prove that the equation to its locus is
, where b2 = a2 (e2 – 1). Solution: Let P (h, k) represent any point on the locus And let the coordinates of A and B be given by (ae, 0) and(-ae, 0). Where, we have: PA – PB = 2a Upon squaring both...
Find the equation of the locus of a point which moves such that the ratio of its distance from (2, 0) and (1, 3) is 5: 4.
Solution: Let P (h, k) represent any point on the locus And let the coordinates of A and B be (2, 0) and (1, 3) respectively. So then, PA/ BP = 5/4 Thus, we can say: PA2 = BP2 =...
Find the locus of a point equidistant from the point (2, 4) and the y-axis.
Solution: Let P (h, k) represent any point on the locus And let the coordinates of A and B be (2, 4) and (0, k). Then, we have PA = PB Or, we can say: PA2 = PB2