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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1

Question 1.

Find the locus of P, if for all values of a, the co-ordinates of a moving point P is

(i) (9 cos α, 9 sin α)

(ii) (9 cos α, 6 sin α)

Solution:

(i) Let P(h, k) be the moving point.

We are given h = 9 cos α and k = 9 sin α and

∴ locus of the point is x^{2} + y^{2} = 81

(ii) Let P(h , k) be a moving point.

We are given h = 9 cos α and k = 6 sin α

Question 2.

Find the locus of a point P that moves at a constant distance of

(i) Two units from the x-axis

(ii) Three units from the y-axis.

Solution:

(i) Let the point (x, y) be the moving point.

The equation of a line at a distance of 2 units from the x-axis is k = 2

So the locus is y = 2 (i.e.) y – 2 = 0

(ii) Equation of a line at a distance of 3 units from y-axis is h = 3

So the locus is x = 3 (i.e.) x – 3 = 0

Question 3.

If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are x = a cos^{3} θ, y = a sin^{3} θ

Solution:

Question 4.

Find the value of k and b, if the points P (-3, 1) and Q (2, b) lie on the locus of x^{2} – 5x + ky = 0.

Solution:

Given P (-3, 1) lies on the locus of

x^{2} – 5x + ky = 0

∴ (- 3)^{2} – 5 (-3) + k(1) = 0

9 + 15 + k = 0

⇒ k = -24

Also given Q(2 , b) lies on the locus of

x^{2} – 5x + ky = 0

x^{2} – 5x – 24y = 0

∴ (2)^{2} – 5(2) – 24(b) = 0

4 – 10 – 24b = 0 ⇒ – 6 – 24b = 0

⇒ 24b = -6 ⇒ b = \(-\frac{6}{24}\) = \(-\frac{1}{4}\)

Thus k = -24, b = \(-\frac{1}{4}\)

Question 5.

A straight rod of length 8 units slides with its ends A and B always on the x and y-axis respectively. Find the locus of the midpoint of the line segment AB.

Solution:

Let P (h, k) be the moving point A (a, 0) and B (0, b) P is the midpoint of AB.

Question 6.

Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, -1) is equal to 20.

Solution:

Let the given points be A (3, 5) and (1, -1).

Let P (h, k ) be the point such that

PA^{2} + PB^{2} = 20 ………….. (1)

PA^{2} = (3 – h)^{2} + (5 – k)^{2}

PB^{2} = (1 – h)^{2} + (- 1 – k)^{2}

(1) ⇒ (3 – h)^{2} + (5 – k)^{2} + (1 – h)^{2} + (1 + k)^{2} = 20

9 – 6h + h^{2} + 25 – 10k + k^{2} + 1 – 2h + h^{2} + 1 + 2k + k^{2} = 20

2h^{2} + 2k^{2} – 8h – 8k + 36 = 20

2h^{2} + 2k^{2} – 8h – 8k + 16 = 0

h^{2} + k^{2} – 4h – 4k + 8 = 0

The locus of P ( h , k ) is obtained by replacing h by x and k by y

∴ The required locus is x^{2} + y^{2} – 4x – 4y + 8 = 0

Question 7.

Find the equation of the locus of the point P such that the line segment AB, joining the points A (1, -6) and B (4, -2), subtends a right angle at P.

Solution:

Let P (h, k) be the moving point

Given A (1, – 6) and B (4, – 2),

Since ∆APB = 90°, PA^{2} + PB^{2} = AB^{2}

(i.e.) (h – 1)^{2} + (k + 6)^{2} + (h – 4)^{2} + (k + 2)^{2} = (4 – 1)^{2} + (-2 + 6)^{2}

(i.e) h^{2} + 1 – 2h + k^{2} + 36 + 12k + h^{2} + 16 – 8h + k^{2} + 4 + 4k = 3^{2} + 4^{2} = 25

2h^{2} + 2k^{2} -10h + 16k + 57 – 25 = 0

2h^{2} + 2k^{2} – 10h + 16k + 32 = 0

(÷ by 2)h^{2} + k^{2} – 5h + 8k + 16 = 0

So the locus of P is x^{2} + y^{2} – 5x + 8y + 16 = 0

Question 8.

If O is origin and R is a variable point on y^{2} = 4x, then find the equation of the locus of the mid-point of the line segment OR.

Solution:

Let P(h, k) be the moving point

We are given O (0, 0). Let R = (a, b)

Substituting a, b values is y^{2} = 4x

we get (2k)^{2} = 4 (2h)

(i.e) 4k^{2} = 8h

(÷ by 4) k^{2} = 2h

So the locus of P is y^{2} = 2x

Question 9.

Solution:

Question 10.

If P (2, -7) is a given point and Q is a point on 2x^{2} + 9y^{2} = 18, then find the equations of the locus of the mid-point of PQ.

Solution:

P = (2, -7); Let (h, k) be the moving point Q = (a, b)

⇒ a = 2h – 2,

b = 2k + l

Q is a point on 2x^{2} + 9y^{2} = 18 (i.e) (a, b) is on 2x^{2} + 9y^{2} = 18

⇒ 2(2h – 2)^{2} + 9 (2k + 7)^{2} = 18

(i.e) 2 [4h^{2} + 4 – 8h] + 9 [4k^{2} + 49 + 28k] – 18 = 0

(i.e) 8h^{2} + 8 – 16h + 36k^{2} + 441 + 252k – 18 = 0

8h^{2} + 36k^{2} – 16h + 252k + 431 = 0

The locus is 8x^{2} + 36y^{2} – 16x + 252y + 431 = 0

Question 11.

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

Solution:

P = (x, 0), Q = (0, y), R (h, k) be a point on RQ such that PR : RQ = b : a

From the right-angled triangle OQR, OR^{2} + OQ^{2} = QR^{2}

Question 12.

If the points P (6, 2) and Q (-2, 1) and R are the vertices of a ∆PQR and R is the point on the locus y = x^{2} – 3x + 4, then find the equation of the locus of the centroid of ∆PQR.

Solution:

P (6, 2), Q (-2, 1). Let R = (a, b) be a point on y = x^{2} – 3x + 4.

But (a, b) is a point on y = x^{2} – 3x + 4

b = a^{2} – 3a + 4

(i.e) 3k – 3 = (3h – 4)^{2} – 3(3h – 4) + 4

(i.e) 3k – 3 = 9h^{2} + 16 – 24h – 9h + 12 + 4

⇒ 9h^{2} – 24h – 9h + 32 – 3k + 3 = 0

(i.e) 9h^{2} – 33h – 3k + 35 = 0,

Locus of (h, k) is 9x^{2} – 33x – 3y + 35 = 0

Question 13.

If Q is a point on the locus of x^{2} + y^{2} + 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 (h, k) be the moving point O = (0, 0);

Let PQ = (a, b) on x^{2} + y^{2} + 4x – 3y + 7 = 0

Question 14.

Find the points on the locus of points that are 3 units from the x-axis and 5 units from the point (5, 1).

Solution:

A line parallel to the x-axis is of the form y = k.

Here k = 3 ⇒ y = 3

A point on this line is taken as P (a, 3).

The distance of P (a, 3) from (5, 1) is given as 5 units

⇒ (a – 5)^{2} + (3 – 1)^{2} = 5^{2}

a^{2} + 25 – 10a + 9 + 1 – 6 = 25

a^{2} – 10a + 25 + 4 – 25 = 0

a^{2} – 10a + 4 = 0

Question 15.

The sum of the distance of a moving point from the points (4, 0) and (-4, 0) is always 10 units. Find the equation of the locus of the moving point.

Solution:

Let P (h, k) be a moving point

Here A = (4, 0) and B = (-4, 0)

Given PA + PB = 10

### Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Additional Questions

Question 1.

If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point.

Solution:

Let coordinates of a moving point P be (x, y).

Given that the sum of the distances from the axis to the point is always 1.

∴ |x| + |y| = 1 ⇒ x + y = 1

⇒ -x – y = 1 ⇒ x – y = 1

Hence, these equations give us the locus of the point P which is a square.

Question 2.

A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x -12y = 3. The equation of its locus is ……..

Solution:

The given equation of line is 5x – 12y = 3 and the given point is (3, -2).

Let (a, b) be any moving point.

⇒ 13a^{2} + 13b^{2} – 78a + 52b + 169 = 5a – 12b – 3

⇒ 13a^{2} + 13b^{2} – 83a + 64b + 172 = 0

So, the locus of the point is 13x^{2} + 13y^{2} – 83x + 64y + 172 = 0

Question 3.

Find the Locus of the midpoints of the portion of the line x cos θ + y sin θ = p intercepted between the axis.

Solution:

Given the equation of the line is x cos θ + y sin θ = p … (i)

Let C (h, k) be the midpoint of the given line AB where it meets the two-axis at A (a, 0) and B (0, b).

Since (a, 0) lies on eq (i) then “a cos θ + θ = p”

B (0, b) also lies on the eq (i) then 0 + b sin θ = p

Since C (h, k) is the midpoint of AB

Putting the values of a and b is eq (ii) and (iii) we get P

Squaring and adding eq (iv) and (v) we get

Question 4.

Solution:

Here, α is a variable. To find the locus of P (h, k), we have to eliminate α.

From (i), we obtain

Question 5.

Solution: