RD Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Ex 22.2 Solutions
Question - 1 : - Find the locus of a point equidistant from the point (2, 4) and the y-axis.
Answer - 1 : -
Let P (h, k) be any point on the locus and let A (2, 4) and B(0, k).
Then, PA = PB
PA2 = PB2
Question - 2 : - 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.
Answer - 2 : -
Let P (h, k) be any point on the locus and let A (2, 0) and B(1, 3).
So then, PA/ BP = 5/4
PA2 = BP2 = 25/16
Question - 3 : - 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).
Answer - 3 : -
Let P (h, k) be any point on the locus and let A (ae, 0) and B(-ae, 0).
Where, PA – PB = 2a
Now again let us square on both the sides we get,
(eh + a)2 = (h +ae)2 + (k – 0)2
e2h2 + a2 +2aeh = h2 + a2e2 +2aeh + k2
h2 (e2 – 1) – k2 =a2 (e2 –1)
Now let us replace (h, k) with (x, y)
The locus of a point such that the difference of its distancesfrom (ae, 0) and (-ae, 0) is 2a.
Where b2 = a2 (e2 –1)
Hence proved.
Question - 4 : - Find the locus of a point such that the sum of its distances from (0, 2) and (0, -2) is 6.
Answer - 4 : -
Let P (h, k) be any point on the locus and let A (0, 2) and B (0, -2).
Where, PA – PB = 6
Question - 5 : - Find the locus of a point which is equidistant from (1, 3) and x-axis.
Answer - 5 : -
Let P (h, k) be any point on the locus and let A (1, 3) and B (h, 0).
Where, PA = PB
Question - 6 : - Find the locus of a point which moves such that its distance from the origin is three times is distance from x-axis.
Answer - 6 : -
Let P (h, k) be any point on the locus and let A (0, 0) and B (h, 0).
Where, PA = 3PB
Now by squaring on both the sides we get,
h2 + k2 = 9k2
h2 = 8k2
By replacing (h, k) with (x, y)
∴ Thelocus of point is x2 = 8y2