RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates Ex-22.1 |
RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates Ex-22.3 |

**Answer
1** :

Let P (h, k) be any point on the locus and let A (2, 4) and B(0, k).

Then, PA = PB

PA^{2} = PB^{2}

**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

PA^{2} = BP^{2} = 25/16

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}

e^{2}h^{2} + a^{2} +2aeh = h^{2} + a^{2}e^{2} +2aeh + k^{2}

h^{2} (e^{2} – 1) – k^{2} =a^{2} (e^{2} –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 b^{2} = a^{2 }(e^{2} –1)

Hence proved.

**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

**Answer
5** :

Let P (h, k) be any point on the locus and let A (1, 3) and B (h, 0).

Where, PA = PB

**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,

h^{2} + k^{2} = 9k^{2}

h^{2} = 8k^{2}

By replacing (h, k) with (x, y)

∴ Thelocus of point is x^{2} = 8y^{2}

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