Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Text Book Back Questions and Answers, Notes.

## 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 coordinates of a moving point P is

(i) (9 cos α, 9 sin α)

(ii) (9 cos α, 6 sin α)

Answer:

(i) (9 cos α, 9 sin α)

Let P (h , k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 9 sin α

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

∴ The required locus becomes x^{2} + y^{2} = 81

(ii) ( 9 cos α, 6 sin α)

Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 6 sin α

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

∴ The required locus is \(\frac{x^{2}}{81}+\frac{y^{2}}{36}\) = 1

Question 2.

Find the locus of a point P that moves a constant distant of

(i) two units from the x-axis

(ii) three units from the y-axis.

Answer:

(i) Two units from x-axis:

Let P (h, k) be any point on the required path. From the given data, we have k = 2

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

∴ The required locus is y = 2

(ii) Three units from y-axis:

Let (h, k) be any point on the required path. From the given data, we have h = 3.

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

∴ The required locus is x = 3

Question 3.

If 6 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.

Answer:

The given moving points is (a cos^{3}θ, a sin^{3}θ)

Question 4.

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

Answer:

Given P (-3, 1) lie on x^{2} – 5x + ky = 0

⇒ (-3)^{2} – 5(-3) + k(1) = 0

9 + 15 + k = 0 ⇒ k = -24

Q (2, b) lie on x^{2} – 5x + ky = 0

(2)^{2} – 5(2) + k(b) = 0 ⇒ 4 – 5(2) – 24b = 0

Question 5.

A straight rod of length 8 units slides with its ends A and B always on the x and y axes respectively, then find the locus of the midpoint of the line segment AB.

Answer:

Given A and B are the ends of the straight rod of length 8 unit on the x and y-axes. Let A be (a, 0) and B (0, b).

Let M (h, k) be the midpoint of AB (h,k) =

In the right-angled ∆ OAB

AB^{2} = OA^{2} + OB^{2}

8^{2} = a^{2} + b^{2}

64 = (2h)^{2} + (2k)

64 = 4h^{2} + 4k^{2}

h^{2} + k^{2} = \(\frac{64}{4}\) = 16

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

∴ The required locus is x^{2} + y^{2} = 16

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.

Answer:

Let P (h, k) be the moving point

Let the given point be A (3, 5) and B (1, -1)

We are given PA^{2} + PB^{2} = 20

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

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

(i.e.) 2h^{2} + 2k^{2} – 8h – 8k + 36 – 20 = 0

2h^{2} + 2k^{2} – 8h – 8k + 16 = 0

(÷ by 2 ) h^{2} + k^{2} – 4h – 4k + 8 = 0

So the locus of P 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.

Answer:

Given A (1, – 6) and B (4, – 2).

Let P (h, k) be a point such that the line segment AB subtends a right angle at P.

∴ ∆ APB is a right-angled triangle.

AB^{2} = AP^{2} + BP^{2} ………… (1)

AB^{2} = (4 – 1)^{2} + (- 2 + 6)^{2}

AB^{2} = 3^{2} + 4^{2} = 9 + 16 = 25

AP^{2} = (h – 1)^{2} + (k + 6)^{2}

BP^{2} = (h – 4)^{2} + (k + 2)^{2}

(1) ⇒

25 = (h – 1)^{2} + (k + 6)^{2} + (h – 4)^{2} + (k + 2)^{2}

25 = h^{2} – 2h + 1 + k^{2} + 12k + 36 + h^{2} – 8h + 16 + k^{2} + 4k + 4

25 = 2h^{2} + 2k^{2} – 10h + 16k + 57

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

2h^{2} + 2k^{2} – 10h + 16k + 32 = 0

h^{2} + k^{2} – 5h + 8k + 16 =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} – 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 segment OR.

Answer:

Let the variable point R be (x, y). Let M (h, k) be the midpoint of R.

But R(x , y) is a point on y^{2} = 4x

∴ (2k)^{2} = 4(2h)

4k^{2} = 8h

k^{2} = 2h

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

∴ The required locus is y^{2} = 2x

Question 9.

The coordinates of a moving point P are (\(\frac{\mathbf{a}}{2}\) (cosec θ + sin θ), \(\frac{\mathbf{b}}{2}\) (cosec θ – sin θ) where θ is a variable parameter. Show that the equation of the locus P is b^{2} x^{2} – a^{2} y^{2} = a^{2} b^{2}

Answer:

Let the moving point P be (h, k)

By the given data we have

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

∴ The required locus is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

b^{2} x^{2} – a^{2} y^{2} = a^{2} b^{2}

Question 10.

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

Answer:

Given P is (2, -7) and let Q be (x, y)

Given that Q is a point on 2x^{2} + 9y^{2} = 18

Let M (h, k) be the midpoint of PQ

2h = 2 + x , 2k = – 7 + y

x = 2h – 2, y = 2k + 7

But Q(x, y) is a point on 2x^{2} + 9y^{2} = 18

∴ 2 (2h – 2)^{2} + 9 (2k + 7)^{2} = 18

2 [4h^{2} – 8h + 4] + 9 [4k^{2} + 28k + 49] = 18

8h^{2} – 16h + 8 + 36k^{2} + 252k + 441 = 18

8h^{2} + 36k^{2} – 16h + 252k + 449 = 18

8h^{2} + 36k^{2} – 16h + 252k +431 =0

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

∴ The required 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 P is a variable point on RQ with RP = b, PQ = a, then find the equation of locus of P.

Answer:

Given R is any point on the x-axis and Q is any point on the y-axis.

Let R be (x, 0) and Q be (0, y)

Let P (h, k ) be the variable point on RQ such that RP = b and PQ = a

The point P ( h, k ) divides the line joining the points R(x, 0) and Q (0, y) in the ratio b : a

Substituting in equation (1), we have

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

∴ The required locus is \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1

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.

Answer:

Given P (6, 2), Q (-2 , 1), R (a, b) are the vertices of ∆ PQR where R (a, b) lies on y = x^{2} – 3x + 4

∴ b = a^{2} – 3a + 4 (1)

Let the centroid of ∆ PQR be G (h, k)

Substituting in equation (1) we have

(1) ⇒ 3k – 3 = (3h – 4)^{2} – 3(3h – 4) + 4

3k – 3 = 9h^{2} – 24h + 16 – 9h + 12 + 4

9h^{2} – 33h + 32 – 3k + 3 = 0

9h^{2} – 33h – 3k + 35 = 0

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

∴ The required locus 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.

Answer:

Let Q be (a, b) lying on the locus

x^{2} + y^{2} + 4x – 3y + 7 = 0

∴ a^{2} + b^{2} + 4a – 3b + 7 = 0

Let the movable point P be (h , k)

Given P divides OQ externally in the ratio 3 : 4

Substituting in equation (1) we have

h^{2} + k^{2} – 12h + 9k + 63 = 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} – 12x + 9y + 63 = 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)

Answer:

Given that the required pointis3unitsfrom x-axis and 5 units from the point P (5, 1). Let Q (h, 3) and K (h,- 3) be the required points.

∴ PQ = 5

\(\sqrt{(5-\mathrm{h})^{2}+(1-3)^{2}}\) = 5

(5 – h)^{2} + (- 2)^{2} = 25

25 – 10h + h^{2} + 4 = 25

h^{2} – 10h + 29 – 25 = 0

h^{2} – 10h + 4 = 0

PR = 5

(5 – k)^{2} + 4^{2} = 25

25 – 10k + k^{2} + 16 = 25

k^{2} – 10k + 16 = 0

∴ R (8, – 3), (2, – 3)

∴ Required points are

(5 + √21, 3), (5 – √21, 3), (8, – 3), (2, – 3)

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 to the locus of the moving point.

Answer:

Let A be (4, 0) and B be (-4, 0). Let the moving point be p(h, k)

Given PA + PB = 10

16h^{2} + 200h + 625 = 25[h^{2} + 8h + 16 + k^{2}]

16h^{2} + 200h + 625 = 25h^{2} + 200h + 400 + 25k^{2}

9h^{2} + 26k^{2} = 225