Inter 1st Year Maths 1B Locus Important Questions

   

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., x2 + 4x + 4 + y2 – 6y + 9 = 25
i.e., x2 + y2 + 4x – 6y – 12 = 0 …………….. (2)
Let Q(x1, y1) satisfy (2).
Then, x12 + y12 + 4x1 – 6y1 – 12 = 0 …………. (3)
Now the distance of A from Q is
AQ = \(\sqrt{\left(x_{1}+2\right)^{2}+\left(y_{1}-3\right)^{2}}\)
∴ AQ2 = x12 + 4x1 + 4 + y12 – 6y1 + 9
= (x12 + y12 + 4x1 – 6y1 – 12) + 25
= 25 (by using (3))
Hence AQ = 5.
This means that Q(x1, y1) satisfies the geometric condition (1).
∴ The required equation of locus is
x2 + y2 + 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., PA2 = 4PB2
i.e., (x – 3)2 + y2 = 4[(x + 3)2 + y2]
i.e., x2 – 6x + 9 + y2 = 4(x2 + 6x + 9 + y2]
i.e., 3x2 + 3y2 + 30x + 27 = 0
i.e., x2 + y2 + 10x + 9 = 0 ………………. (2)
i.e., Q(x1, y1) satisfy (2).
Then x12 + y12 + 10x1 + 9 = 0 …………….. (3)
Now QA2 = (x1 – 3)2 + y12
= x12 – 6x1 + 9 + y12
= 4x12 + 24x1 + 36 + 4y12 – 3x12 – 30x1 – 27 – 3y12
= 4(x12 + 6x1 + 9 + y12) – 3(x12 + 10x1 + 9 + y12)
= 4(x12 + 6x1 + 9 + y12) (by using(3))
= 4 [(x1 + 3)2 + y12]
= 4QB2
∴ QA = 2QB.
This means that Q(x1, y1) satisfies (1).
Hence, the required equation of locus is
x2 + y2 + 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 PA2 + PB2 = AB2.
i.e., (x – 4)2 + y2 + x2 + (y – 4)2 = 16 + 16
i.e., 2x2 + 2y2 – 8x – 8y = 0
or x2 + y2 – 4x – 4y = 0
Let Q(x1, y1) satisfy (2) and Q be different from A and B.
Then x12 + y12 — 4x1 – 4y1 = 0,
(x1, y1) ≠ (4, 0) and (x1, y1) ≠ (0, 4) ……………… (3)
Now QA2 + QB2 = (x1 – 4)2 + y12 + x12 + (y1 – 4)2
= x12 – 8x1 + 16 + y12 + x12 + y12 – 8y1 + 16
= 2(x12 + y12 – 4x1 – 4y1) + 32
= 32 (by using (3))
= AB2
Hence QA2 + QB2 = AB2, Q ≠ A and Q ≠ B.
This means that Q(x1, y1) 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., 9AP2 = 4PB2
i.e., 9[(x – 5)2 + (y + 4)2] = 4[(x – 7)2 + (y – 6)2]
i.e., 9[x2 + 25 – 10x + y2 + 16 + 8y]
= 4[x2 + 49 – 14x + y2 + 36 – 12y]
i.e., 5x2 + 59 – 34x + 120y + 29 = 0 ……………….. (2)
Let Q(x1, y1) satisfy (2). Then
5x12 + 5y12 – 34x1 + 120y1 + 29 = 0 ……………….. (3)
Now 9AQ2 = 9[x12 + 25 – 10x1 + y12 + 16 + 8y1]
= 5x12 + 5y12 – 34x1 + 120y1 + 29 + 4x12 + 4y12 – 56x1 – 48y1 + 340
= 4[x12 + y12 – 14x1 – 12y1 + 49 + 36] (by using (3))
= 4[(x1 – 7)2 + (y1 – 6)2] = 4PB2
Thus 3AQ = 2PB. This means that Q(x1, y1) satisfies (1).
Hence, the required equation of locus is
5(x2 + y2) – 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., x2 + 10xy + 25y2 – 34x – 170y = 0 …………… (2)
Let Q(x1, y1) satisfy (2). Then
x1 + 5y1 = 0 or x1 + 5y1 = 34 ……………………. (3)
Now, area of ∆QAB
= \(\frac{1}{2}\)|x1(3 – 4) + 2(4 – y1) – 3(y1 – 3)|
= \(\frac{1}{2}\)|-x1 + 8 – 2y1 – 3y1 + 9|
= \(\frac{1}{2}\) |-x1 – 5y1 + 17|
= \(\frac{17}{2}\) = 8.5 (by using (3))
This means that Q(x1, y1) satisfies (1).
Hence, the required equation of locus is
(x + 5y)(x + 5y – 34) = 0 or
x2 + 10xy + 25y2 – 34x – 170y = 0.