📘 Locus – Maths 1B | EAPCET PYQs (2019–2024) with Full Solutions
Subject: Mathematics 1B
Exam: TS & AP EAPCET
Prepared by: Aimstutorial
Covers: 2019–2024 PYQs + step-by-step solutions
📗 Important Formulas
| Condition | Equation | Curve Type |
|---|---|---|
| Distance from fixed point is constant | \\[(x-h)^2 + (y-k)^2 = r^2\\] | Circle |
| Sum of distances from two fixed points is constant | \\[PA + PB = k\\] | Ellipse |
| Difference of distances from two fixed points is constant | \\[|PA – PB| = k\\] | Hyperbola |
| Distance from point equals distance from line | \\[r = \\dfrac{|ax + by + c|}{\\sqrt{a^2 + b^2}}\\] | Parabola |
| Equal distance from two points | \\[PA = PB\\] | Perpendicular bisector (line) |
🎯 EAPCET PYQs (Snapshot)
| Year | Question Summary | Equation / Answer | Exam |
|---|---|---|---|
| 2020 | Locus with distance ratio from O and (−2,−3) = 5:7 | \\[24x^2+24y^2-100x-150y-325=0\\] | AP EAMCET 21-09-2020 S1 |
| 2020 | PA = PB with A(2,3), B(−2,1) | \\[2x + y = 2\\] | AP/TS |
| 2022 | \\(PA^2 + PB^2 = PC^2\\) type | \\[x^2 + y^2 – 6x – 2y + 2 = 0\\] | TS EAMCET 19-07-2022 S2 |
| 2023 | Equidistant from (1,1) and line x+y+1=0 | \\[x^2 + y^2 – 6x + 4y – 3 = 0\\] | TS EAMCET 17-05-2023 S1 |
🧩 20 PYQs with Full, Collapsible Solutions
Find the locus of P such that ratio of distances from O(0,0) and A(−2,−3) is 5:7.
Solution
Let P(x,y). \\(\\dfrac{OP}{AP}=\\dfrac{5}{7}\\Rightarrow \\dfrac{\\sqrt{x^2+y^2}}{\\sqrt{(x+2)^2+(y+3)^2}}=\\dfrac{5}{7}\\). Squaring:
\\[49(x^2+y^2)=25\\{(x+2)^2+(y+3)^2\\}\\]P is equidistant from A(2,3) and B(−2,1). Find the locus.
Solution
\\[(x-2)^2+(y-3)^2=(x+2)^2+(y-1)^2\\]For A(2,0), B(−2,0), the sum \\(PA+PB=6\\). Find the locus.
Solution
Ellipse with foci at \\(\\pm2\\) on x-axis. \\(2a=6\\Rightarrow a=3,\\;c=2\\Rightarrow b^2=a^2-c^2=5\\).
For A(3,0), B(−3,0), the difference \\(|PA-PB|=2\\). Find the locus.
Solution
Hyperbola with foci \\(\\pm3\\). \\(2a=2\\Rightarrow a=1,\\;c=3\\Rightarrow b^2=c^2-a^2=8\\).
P is equidistant from S(1,1) and the line \\(x+y+1=0\\). Find the locus.
Solution
\\[\\sqrt{(x-1)^2+(y-1)^2}=\\frac{|x+y+1|}{\\sqrt2}\\]Given A(0,0), B(4,0). Find locus of P such that \\(PA=2\\,PB\\).
Solution
\\[\\sqrt{x^2+y^2}=2\\sqrt{(x-4)^2+y^2}\\Rightarrow 3x^2+3y^2-32x+64=0\\]A rod of length \\(l\\) has ends on axes. Find locus of its midpoint.
Solution
Endpoints \\((a,0),(0,b)\\) with \\(a^2+b^2=l^2\\). Midpoint \\(M(\\tfrac a2,\\tfrac b2)=(x,y)\\).
“Sum of squares of distances from axes” equals “square of distance from line \\(x-y-1=0\\)”.
Solution
LHS \\(=x^2+y^2\\), RHS \\(=\\dfrac{(x-y-1)^2}{2}\\).
With A(1,1), B(2,3), area of \\(\\triangle PAB=9\\). Find locus of P.
Solution
Using area formula → \\(|y+1-2x|=18\\).
For A(0,a), B(0,−a), find locus of points P with \\(PA=PB\\).
Solution
The sum of distances of P from the axes is 1. Describe the locus.
Solution
Triangle with vertices (0,0),(a,0),(0,b) has centroid G. If \\(a+b=6\\), find locus of G.
Solution
\\(G(\\tfrac a3,\\tfrac b3)=(x,y)\\Rightarrow 3x+3y=6\\).
Distance of P from (2,0) equals twice its distance from y-axis. Find locus.
Solution
\\[\\sqrt{(x-2)^2+y^2}=2|x|\\Rightarrow (x-2)^2+y^2=4x^2\\]The product of distances of P from the axes is \\(k>0\\). Find locus.
Solution
Tangents drawn from P to \\(x^2+y^2=r^2\\) have length \\(l\\). Find locus.
Solution
Power of a point: \\(PT^2=OP^2-r^2=l^2\\Rightarrow x^2+y^2=r^2+l^2\\).
Midpoints of chords of \\(x^2+y^2=a^2\\) that are parallel to x-axis. Find locus.
Solution
For fixed A(c,0), B(−c,0), the quantity \\(PA^2+PB^2=m^2\\) is constant.
Solution
\\[(x-c)^2+y^2+(x+c)^2+y^2=2(x^2+y^2+c^2)=m^2\\]Distance of P from line \\(ax+by+c=0\\) is a fixed \\(k\\). Find locus.
Solution
If \\(\\angle APB=90^\\circ\\) where A(−2,0), B(2,0), find locus of P.
Solution
Thale’s theorem → circle with AB as diameter.
The sum of distances of P from the axes is constant \\(k>0\\). Find locus and type.
Solution
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