Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Locus Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1B Locus Important Questions

Question 1.

Find the equation of the locus of a point which is at a distance 5 from (-2, 3) in the xoy plane.

Solution:

Let the given point be A = (-2, 3) and P(x, y) be a point on the plane.

The geometric condition to be satisfied by P to be on the locus is that

AP = 5 …………… (1)

Expressing this condition algebraically, we get

\(\sqrt{(x+2)^{2}+(y-3)^{2}}\) = 5

i.e., x^{2} + 4x + 4 + y^{2} – 6y + 9 = 25

i.e., x^{2} + y^{2} + 4x – 6y – 12 = 0 …………….. (2)

Let Q(x_{1}, y_{1}) satisfy (2).

Then, x_{1}^{2} + y_{1}^{2} + 4x_{1} – 6y_{1} – 12 = 0 …………. (3)

Now the distance of A from Q is

AQ = \(\sqrt{\left(x_{1}+2\right)^{2}+\left(y_{1}-3\right)^{2}}\)

∴ AQ^{2} = x_{1}^{2} + 4x_{1} + 4 + y_{1}^{2} – 6y_{1} + 9

= (x_{1}^{2} + y_{1}^{2} + 4x_{1} – 6y_{1} – 12) + 25

= 25 (by using (3))

Hence AQ = 5.

This means that Q(x_{1}, y_{1}) satisfies the geometric condition (1).

∴ The required equation of locus is

x^{2} + y^{2} + 4x – 6y – 12 = 0.

Question 2.

Find the equation of locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).

Solution:

Let P(x, y) he a point on the locus. Then the geometric condition to be satisfied by P is

PA = 2PB …………….. (1)

i.e., PA^{2} = 4PB^{2}

i.e., (x – 3)^{2} + y^{2} = 4[(x + 3)^{2} + y^{2}]

i.e., x^{2} – 6x + 9 + y^{2} = 4(x^{2} + 6x + 9 + y^{2}]

i.e., 3x^{2} + 3y^{2} + 30x + 27 = 0

i.e., x^{2} + y^{2} + 10x + 9 = 0 ………………. (2)

i.e., Q(x_{1}, y_{1}) satisfy (2).

Then x_{1}^{2} + y_{1}^{2} + 10x_{1} + 9 = 0 …………….. (3)

Now QA^{2} = (x_{1} – 3)^{2} + y_{1}^{2}

= x_{1}^{2} – 6x_{1} + 9 + y_{1}^{2}

= 4x_{1}^{2} + 24x_{1} + 36 + 4y_{1}^{2} – 3x_{1}^{2} – 30x_{1} – 27 – 3y_{1}^{2}

= 4(x_{1}^{2} + 6x_{1} + 9 + y_{1}^{2}) – 3(x_{1}^{2} + 10x_{1} + 9 + y_{1}^{2})

= 4(x_{1}^{2} + 6x_{1} + 9 + y_{1}^{2}) (by using(3))

= 4 [(x_{1} + 3)^{2} + y_{1}^{2}]

= 4QB^{2}

∴ QA = 2QB.

This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

x^{2} + y^{2} + 10x + 9 = 0.

Question 3.

Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).

Solution:

Let A = (4, 0) and B = (0, 4).

Let P(x, y) be a point such that. PA and PB are perpendicular. Then PA^{2} + PB^{2} = AB^{2}.

i.e., (x – 4)^{2} + y^{2} + x^{2} + (y – 4)^{2} = 16 + 16

i.e., 2x^{2} + 2y^{2} – 8x – 8y = 0

or x^{2} + y^{2} – 4x – 4y = 0

Let Q(x_{1}, y_{1}) satisfy (2) and Q be different from A and B.

Then x_{1}^{2} + y_{1}^{2} — 4x_{1} – 4y_{1} = 0,

(x_{1}, y_{1}) ≠ (4, 0) and (x_{1}, y_{1}) ≠ (0, 4) ……………… (3)

Now QA^{2} + QB^{2} = (x_{1} – 4)^{2} + y_{1}^{2} + x_{1}^{2} + (y_{1} – 4)^{2}

= x_{1}^{2} – 8x_{1} + 16 + y_{1}^{2} + x_{1}^{2} + y_{1}^{2} – 8y_{1} + 16

= 2(x_{1}^{2} + y_{1}^{2} – 4x_{1} – 4y_{1}) + 32

= 32 (by using (3))

= AB^{2}

Hence QA^{2} + QB^{2} = AB^{2}, Q ≠ A and Q ≠ B.

This means that Q(x_{1}, y_{1}) satisfies (1).

∴ The required equation of locus is (2),which is the circle with \(\overline{\mathrm{AB}}\) as diameter, deleting the points A and B.

Though A and B satisfy equation (2), they do not satisfy the required geometric condition.

Question 4.

Find the equation of the locus of P, if the ratio of the distances from P to A(5, -4) and B(7, 6) is 2 : 3. [Mar 14]

Solution:

Let P(x, y) be any point on the locus.

The geometric condition to be satisfied by P

is \(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{1}{2}\)

i.e., 3AP = 2PB …………….. (1)

i.e., 9AP^{2} = 4PB^{2}

i.e., 9[(x – 5)^{2} + (y + 4)^{2}] = 4[(x – 7)^{2} + (y – 6)^{2}]

i.e., 9[x^{2} + 25 – 10x + y^{2} + 16 + 8y]

= 4[x^{2} + 49 – 14x + y^{2} + 36 – 12y]

i.e., 5x^{2} + 59 – 34x + 120y + 29 = 0 ……………….. (2)

Let Q(x_{1}, y_{1}) satisfy (2). Then

5x_{1}^{2} + 5y_{1}^{2} – 34x_{1} + 120y_{1} + 29 = 0 ……………….. (3)

Now 9AQ^{2} = 9[x_{1}^{2} + 25 – 10x_{1} + y_{1}^{2} + 16 + 8y_{1}]

= 5x_{1}^{2} + 5y_{1}^{2} – 34x_{1} + 120y_{1} + 29 + 4x_{1}^{2} + 4y_{1}^{2} – 56x_{1} – 48y_{1} + 340

= 4[x_{1}^{2} + y_{1}^{2} – 14x_{1} – 12y_{1} + 49 + 36] (by using (3))

= 4[(x_{1} – 7)^{2} + (y_{1} – 6)^{2}] = 4PB^{2}

Thus 3AQ = 2PB. This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

5(x^{2} + y^{2}) – 34x + 120y + 29 = 0.

Question 5.

A(2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5. [Mar 11]

Solution:

Let P(x, y) be a point on the locus.

The geometric condition to be satisfied by P is that,

area of ∆PAB = 8.5 …………………(1)

i.e., \(\frac{1}{2}\)|x(3 – 4) + 2(4 – y) – 3(y – 3)| = 8.5

i.e., |-x + 8 – 2y – 3y + 9| = 17

i.e., |-x – 5y + 17| = 17

i.e., -x – 5y + 17 = 17 or -x – 5y + 17 = -17

i.e., x + 5y = 0 or x + 5y = 34

∴ (x + 5y) (x + 5y – 34) = 0

i.e., x^{2} + 10xy + 25y^{2} – 34x – 170y = 0 …………… (2)

Let Q(x_{1}, y_{1}) satisfy (2). Then

x_{1} + 5y_{1} = 0 or x_{1} + 5y_{1} = 34 ……………………. (3)

Now, area of ∆QAB

= \(\frac{1}{2}\)|x_{1}(3 – 4) + 2(4 – y_{1}) – 3(y_{1} – 3)|

= \(\frac{1}{2}\)|-x_{1} + 8 – 2y_{1} – 3y_{1} + 9|

= \(\frac{1}{2}\) |-x_{1} – 5y_{1} + 17|

= \(\frac{17}{2}\) = 8.5 (by using (3))

This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

(x + 5y)(x + 5y – 34) = 0 or

x^{2} + 10xy + 25y^{2} – 34x – 170y = 0.