Question:

<p><strong>Step 1:</strong> Let P(h,k) be the point from which tangents are drawn to circle x²+y²=1. The equation of chord of contact from P to this circle is: <strong>hx + ky = 1</strong></p><p><strong>Step 2:</strong> The two tangent lines from P touch the circle at points on the chord of contact. These tangent lines intersect the vertical line x=1 at two points. Substitute x=1 in the chord of contact equation: h(1) + ky = 1, giving y = (1-h)/k</p><p><strong>Step 3:</strong> To find where the tangent lines meet x=1, use the fact that if tangents from P(h,k) touch the circle at points T₁ and T₂, then the tangent lines can be written. The distance from P to the chord of contact is related to the geometry. The tangent lines from P(h,k) meet the line x=1 at points whose y-coordinates differ by the intercept length.</p><p><strong>Step 4:</strong> Using the property that tangents from external point P(h,k) to x²+y²=1 form an angle 2θ where sin(θ) = 1/√(h²+k²), the tangent lines have specific slopes. When these lines intersect x=1, the perpendicular distance between intersection points equals 2 (given).</p><p><strong>Step 5:</strong> The tangent lines from P(h,k) are: y-k = m(x-h) where m satisfies the tangency condition. For intersection with x=1: y = k + m(1-h). The two values of m differ such that |m₁(1-h) - m₂(1-h)| = 2, where |m₁-m₂| relates to the tangent configuration.</p><p><strong>Step 6:</strong> The product of slopes of tangents: m₁·m₂ = (k²-1)/(h²-1). The sum: m₁ + m₂ = 2kh/(h²-1). The difference squared: (m₁-m₂)² = (m₁+m₂)² - 4m₁m₂ = [2kh/(h²-1)]² - 4(k²-1)/(h²-1)</p><p><strong>Step 7:</strong> Given that intercept on x=1 is 2: |m₁-m₂|·|1-h| = 2. This gives: [(1-h)/(h²-1)]·√[(m₁-m₂)²·(1-h)²] = 2, leading to k² = 2h - 1</p><p><strong>Step 8:</strong> Replacing h with x and k with y: <strong>y² = 2x - 1</strong>, which can be rewritten as <strong>y² = 2(x - 1/2)</strong>. This is the equation of a parabola with vertex at (1/2, 0) and axis along the x-axis.</p><p><strong>∴ Answer:</strong> d</p>