Question 16

(P) Circle centre (1,2), radius = |3+8−1|/5 = 2. Point (3,2): distance from centre = 2 ✓. So P→3. (Q) Tangent to y²=8x: y = mx + 2/m. Distance from origin = √2: (2/m)² = 2(1+m²) → 4/m² = 2+2m² → 2m⁴+2m²−4=0 → m²=1, m=1. Line y = x+2 passes through (7,9) ✓. So Q→2. (R) Ellipse x²/16 + y²/12 = 1, a=4, b=2√3, c=2. M = (2, 3). Normal: (x−2)/(2/16) = (y−3)/(3/12) → (x−2)·8 = (y−3)·4 → 2(x−2) = y−3 → y = 2x−1. Point (1,1) ✓. So R→1. (S) Hyperbola with focus (5,0), directrix x = −16/5 (corresponding to focus (−5,0)): a/e = 16/5, ae = 5 → a² = 16, a=4, e=5/4, b²=9. x²/16 − y²/9 = 1. Point (8, 3√3): 64/16 − 27/9 = 4−3 = 1 ✓. So S→5. Answer: B

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**Solving for tangents** I'm calculating the radius distance from the center to the line and find it to be 2. I've checked several points against the circle equation and only one passes, which is interesting because it narrows down my options. Then, for the tangent line question, I need a line that meets both the circle and the parabola with a positive slope. I’m setting up y = mx + c, while using the parabola's known parameters. It feels like I'm piecing together a puzzle here! **Deriving tangent slope** I'm looking at the equation for a conic section and wondering if the formula I have is correct. It seems like I'm trying to derive the tangent slope for given values. If I plug in x1=2, y1=3, a²=16, and b²=12, the slope calculates to -1/2. This looks like a reasonable result, but I need to double-check the formulas and ensure everything aligns correctly. Let’s keep analyzing! **Exploring alternatives** I need to check for alternative interpretations and ensure a robust solution. My final output should be in JSON format. There’s answer B, but it seems unanswered or marked false. I want to consider the image provided and the options given. For the question regarding a common tangent with positive slope, it seems like y=x+2 is the candidate. But then, would a line like y=x-2 have a positive slope too? I wonder about tangents to the parabola.

For part P, I'm finding the circle's radius using the distance from center (1,2) to the given line, getting radius 2, so the circle equation is (x-1)²+(y-2)²=4. Now I'm checking which points satisfy this equation, starting with (1,1) which doesn't work. For part Q, I'm working with the circle x²+y²=2 and parabola y²=8x, finding the common tangent with positive slope using the tangent line formula for parabolas, solving for m and getting m=1 with the tangent line y=x+2. I'm now checking which points lie on this line, and (7,9) satisfies it. Since Q maps to point 2, this narrows things down to options B and C, since options A and D have different Q mappings. I'm now moving on to part R, working with the ellipse 3x²+4y²=48, converting it to standard form to get a²=16 and b²=12, finding c=2 and the latus rectum endpoint M=(2,3) in the first quadrant, then setting up the normal line equation at that point using the standard ellipse normal formula. Simplifying gives 2x-y=1, and testing points I find (1,1) satisfies this, so R maps to point 1. This confirms option B since it's consistent with both Q→2 and R→1, while option C fails since it requires R→4. Now I need to verify S for option B using the hyperbola problem with focus at (5,0) and directrix x=-16/5, checking whether point (8,3√3) satisfies the hyperbola's defining property via the focus-directrix relationship. So the hyperbola equation becomes x²/16 - y²/9=1. Checking the point (8,3√3): 64/16 - 27/9 = 4-3=1, which confirms it works, so S=5 matches option B. Looking at each part: **Part P:** Circle center (1,2) touching line 3x+4y=1. Radius = |3+8-1|/5 = 2, so circle: (x-1)²+(y-2)²=4. Check (3,2): 4+0=4 ✓ → **P→3** **Part Q:** Common tangent to x²+y²=2 and y²=8x (a=2) with positive slope. Tangent to parabola: y = mx + 2/m. Tangent condition to circle: (2/m)²/(1+m²) = 2 → 4 = 2m²(1+m²) → m⁴+m²-2=0 → m²=1, m=1. Line: y = x+2. Check (7,9): 9 = 7+2 = 9 ✓ → **Q→2** **Part R:** Ellipse 3x²+4y²=48 → x²/16+y²/12=1, a²=16, b²=12, c²=4, c=2. Latus rectum endpoint (first quadrant): x=2, y=b²/a=3 → M=(2,3). Normal: a²x/x₀ - b²y/y₀ = a²-b² → 8x - 4y = 4 → 2x - y = 1. Check (1,1): 2-1=1 ✓ → **R→1** **Part S:** Hyperbola with focus (5,0), directrix 5x+16=0 → x=-16/5. Since directrix is negative but focus positive, it corresponds to other focus: a²/c = 16/5, c=5 → a²=16, b²=c²-a²=9. Hyperbola: x²/16 - y²/9 = 1. Check (8, 3√3): 64/16 - 27/9 = 4-3 = 1 ✓ → **S→5** All matches confirm: P→3, Q→2, R→1, S→5, which is option **B**. ```json