LOCUS OF A POINT EXAMPLES

About "Locus of a Point Examples"

Locus of a point examples :

Here we are going to see how to find equation of locus of a point with the given condition.

locus of a point examples - Questions

Question 1 :

If R is any point on the x-axis and Q is any point on the y-axis and P is a variable point on RQ with RP = b, PQ = a. then find the equation of locus of P.

Solution :

QR  =  a + b

In triangle PAR

sin θ  =  AP/PR

sin θ  =  k/b  ---(1)

In triangle QBP

cos θ  =  BP/QP

cos θ  =  h/a  ---(2)

sin2θ + cos2θ  = (k/b)2 + (h/a)2

1  =  (k2/b2) + (h2/a2)

(h2/a2) + (k2/b2)  =  1

Question 2 :

If the points P(6, 2) and Q(−2, 1) and R are the vertices of a ΔPQR and R is the point on the locus y = x2 − 3x + 4, then find the equation of the locus of centroid of ΔPQR

Solution :

Let R (a, b) 

Centroid of the triangle PQR be (h, k)

P(6, 2) Q(-2, 1) R(a, b)

Centroid of the triangle  =  (x1+x2+x3)/3, (y1+y2+y3)/3

(h, k)  =  (6-2+a)/3, (2+1+b)/3

(h, k)  =  (4+a)/3, (3+b)/3

h  =  (4+a)/3

3h = 4 + a 

a = 3h - 4

k  =  (3+b)/3

3k = 3 + b

b = 3k - 3

Since the point (a, b) lies on the curve y = x2 − 3x + 4

b = a2 − 3a + 4

3k - 3  =  (3h - 4)2 − 3(3h - 4) + 4

3k - 3  =  9h2 + 16 + 24h + 9h + 12 + 4

9h2 + 33h - 3k + 32 + 3  =  0

9h2 + 33h - 3k + 35  =  0

Question 3 :

If Q is a point on the locus of x2 + y2 + 4x − 3y + 7 = 0, then find the equation of locus of P which divides segment OQ externally in the ratio 3:4, where O is origin.

Solution :

Let Q be (a, b) and P (h, k)

P is the point which divides the line segment joining the point OQ externally in the ratio 3 : 4

O (0, 0) Q (a, b)

  =  (mx2 - mx1)/(m - n), (my2 - my1)/(m - n)

  =  (3a - 0)/(3-4), (3b - 0)/(3-4)

 (h, k)  =  (-3a, -3b)

-3a  =  h

a  =  -h/3

-3b  =  k

b  =  -k/3

Q is a point on the locus of x2 + y2 + 4x − 3y + 7 = 0

a2 + b2 + 4a − 3b + 7 = 0

By applying the values of "a" and "b" in the above equation, we get

(-h/3)2 + (-k/3)2 + 4(-h/3) - 3(-k/3) + 7 = 0

(h2/9) + (k2/9) - (4h/3) + k + 7  =  0

h2+ k- 12h + 9k + 63  =  0

Here h = x and k = y

x2+ y- 12x + 9y + 63  =  0

After having gone through the stuff given above, we hope that the students would have understood "Locus of a Point Examples". 

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