3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Let S=(1,2) and let L be the line x+y=1. The locus of a point P for which (perpendicular distance of P from L)PS=1 (eccentricity 1) is a parabola P with focus S and directrix L. The pair (vertex of P, length of the latus rectum of P) is ______. Exactly one option is correct.
A(21,23) and 2
B(0,1) and 22
C(21,23) and 22
D(21,23) and 42
Show answer and solution
Answer: C
The axis is the line through S perpendicular to L: x−y+1=0. It meets L at Z=(0,1) (solve x+y=1, x−y=−1). The vertex is the midpoint of SZ: (21,23).
The focus-directrix distance is SZ=2∣1+2−1∣=2. Take new perpendicular axes with origin at the vertex, one along the axis of the parabola (a rotation through 45∘ plus a shift, which preserves lengths); the equation becomes Y2=4aX with 2a=SZ=2, so the latus rectum is 4a=22.
Check: the latus rectum lies on x+y=3 (through S, parallel to L); every point on it is at distance 2 from L, so its ends are the points of x+y=3 at distance 2 from S, giving length 22.
Trap: taking the latus rectum as 4×SZ gives 42; taking Z as the vertex gives (0,1).
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
For the parabola x2−4x−8y+28=0, let S be its focus and let P be a variable point on the parabola. For every position of P, the circle described on SP as diameter touches a fixed line. That line is ______. Exactly one option is correct.
Ay=1
By=3
Cy=5
Dx=2
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Answer: B
Complete the square: (x−2)2=8(y−3). Shift the origin to the vertex: X=x−2, Y=y−3 gives X2=8Y, so a=2, the focus is (X,Y)=(0,2), i.e. S=(2,5), and the directrix is Y=−2, i.e. y=1.
Parametrise: P=(4t,2t2) in XY-coordinates (it satisfies X2=8Y). By the focus-directrix property, SP=YP+2=2t2+2.
Centre of the circle: (2t,22t2+2)=(2t,t2+1); radius =2SP=t2+1.
The centre's distance from Y=0 equals the radius for every t, so the circle always touches Y=0, i.e. the tangent at the vertex y=3.
The other lines fail: distance to y=1 is t2+3>t2+1; distance to y=5 equals ∣t2−1∣, which equals the radius only when t=0; the line x=2 is at distance 2∣t∣, equal to t2+1 only when t=±1.
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
A line L touches both the parabolas y2=4x and x2=32y. The distance between the point where L touches y2=4x and the point where L touches x2=32y is ______. Exactly one option is correct.
A65
B45
C310
D25
Show answer and solution
Answer: A
Non-vertical tangents to y2=4x are y=mx+m1 (m=0), touching at (m21,m2). The line x=0 touches y2=4x but meets x2=32y only at the origin transversally (it is the axis of that parabola), so it is not a common tangent.
Substitute y=mx+m1 in x2=32y: x2−32mx−m32=0. Tangency needs discriminant 0: 1024m2+m128=0⟺m3=−81⟺m=−21 (only real root). So L is unique: y=−2x−2.
Contact with y2=4x: (m21,m2)=(4,−4) (check: 16=4⋅4). Contact with x2=32y: double root x=16m=−8, y=3264=2 (check: on L, −2−8−2=2).
Distance =122+62=180=65.
Difficulty: Very hard
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