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Question: Find Nalure of locus of Pt. Which moves such that its distance from (1,-3) is/double of its distance...

Find Nalure of locus of Pt. Which moves such that its distance from (1,-3) is/double of its distance from 2x-y-5=0

Answer

Pair of intersecting lines

Explanation

Solution

Let P(x, y) be the moving point.
The fixed point is A(1, -3).
The fixed line is L: 2x - y - 5 = 0.

The distance of P from A is PA=(x−1)2+(y−(−3))2=(x−1)2+(y+3)2PA = \sqrt{(x - 1)^2 + (y - (-3))^2} = \sqrt{(x - 1)^2 + (y + 3)^2}.
The distance of P from the line L is PL=∣2x−y−5∣22+(−1)2=∣2x−y−5∣5PL = \frac{|2x - y - 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|2x - y - 5|}{\sqrt{5}}.

According to the problem, the distance from (1, -3) is double the distance from 2x - y - 5 = 0.
So, PA = 2 * PL.
(x−1)2+(y+3)2=2⋅∣2x−y−5∣5\sqrt{(x - 1)^2 + (y + 3)^2} = 2 \cdot \frac{|2x - y - 5|}{\sqrt{5}}.

Square both sides:
(x−1)2+(y+3)2=(2∣2x−y−5∣5)2(x - 1)^2 + (y + 3)^2 = \left( \frac{2 |2x - y - 5|}{\sqrt{5}} \right)^2
(x2−2x+1)+(y2+6y+9)=4(2x−y−5)25(x^2 - 2x + 1) + (y^2 + 6y + 9) = \frac{4 (2x - y - 5)^2}{5}
x2+y2−2x+6y+10=45(4x2+y2+25−4xy−20x+10y)x^2 + y^2 - 2x + 6y + 10 = \frac{4}{5} (4x^2 + y^2 + 25 - 4xy - 20x + 10y)

Multiply by 5:
5(x2+y2−2x+6y+10)=4(4x2+y2+25−4xy−20x+10y)5(x^2 + y^2 - 2x + 6y + 10) = 4(4x^2 + y^2 + 25 - 4xy - 20x + 10y)
5x2+5y2−10x+30y+50=16x2+4y2+100−16xy−80x+40y5x^2 + 5y^2 - 10x + 30y + 50 = 16x^2 + 4y^2 + 100 - 16xy - 80x + 40y

Rearrange the terms to get the general equation of the locus:
0=16x2−5x2+4y2−5y2−16xy−80x+10x+40y−30y+100−500 = 16x^2 - 5x^2 + 4y^2 - 5y^2 - 16xy - 80x + 10x + 40y - 30y + 100 - 50
0=11x2−y2−16xy−70x+10y+500 = 11x^2 - y^2 - 16xy - 70x + 10y + 50
The equation of the locus is 11x2−16xy−y2−70x+10y+50=011x^2 - 16xy - y^2 - 70x + 10y + 50 = 0.

This is a second-degree equation of the form Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A = 11, B = -16, C = -1.
The nature of the conic section is determined by the discriminant B2−4ACB^2 - 4AC.
Discriminant = (−16)2−4(11)(−1)=256+44=300(-16)^2 - 4(11)(-1) = 256 + 44 = 300.

Since B2−4AC=300>0B^2 - 4AC = 300 > 0, the locus is a hyperbola (or a degenerate pair of intersecting lines).

To check for degeneracy, we evaluate the determinant of the matrix associated with the conic:
M=(AB/2D/2B/2CE/2D/2E/2F)=(11−8−35−8−15−35550)M = \begin{pmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{pmatrix} = \begin{pmatrix} 11 & -8 & -35 \\ -8 & -1 & 5 \\ -35 & 5 & 50 \end{pmatrix}
The determinant of M is:
det(M) = 11((−1)(50)−(5)(5))−(−8)((−8)(50)−(5)(−35))+(−35)((−8)(5)−(−1)(−35))11((-1)(50) - (5)(5)) - (-8)((-8)(50) - (5)(-35)) + (-35)((-8)(5) - (-1)(-35))
det(M) = 11(−50−25)+8(−400+175)−35(−40−35)11(-50 - 25) + 8(-400 + 175) - 35(-40 - 35)
det(M) = 11(−75)+8(−225)−35(−75)11(-75) + 8(-225) - 35(-75)
det(M) = −825−1800+2625-825 - 1800 + 2625
det(M) = −2625+2625=0-2625 + 2625 = 0.

Since the determinant is 0 and B2−4AC>0B^2 - 4AC > 0, the conic section is a degenerate hyperbola, which is a pair of intersecting lines.