Question

For the parabola \(y = x^2 - 3x + 2\), match the items in list-1 to that of the items in list-2.
S is a focus, Z is intersection of axis and directrix, P is one end point of latus rectum, Q is the point on the parabola at which tangent is parallel to X-axis

• A. A -- I, B -- II, C -- III, D -- IV
• B. A -- I, B -- II, C -- V, D -- IV
• C. A -- II, B -- V, C -- III, D -- IV
• D. A -- IV, B -- V, C -- III, D -- I

Hint:

For a parabola \(y = ax^2 + bx + c\), the vertex is at \(x = -b/2a\). This is also the point where the tangent is horizontal.

Answer 1

The Correct Option is A

Step 1: Convert Parabola to Standard Form:
Equation: \(y = x^2 - 3x + 2\) Complete the square for \(x\): \[ y = \left(x - \frac{3}{2}\right)^2 + 2 - \frac{9}{4} \] \[ y = \left(x - \frac{3}{2}\right)^2 - \frac{1}{4} \] \[ \left(x - \frac{3}{2}\right)^2 = 1 \cdot \left(y + \frac{1}{4}\right) \] Standard form: \((x - h)^2 = 4a(y - k)\). Here, vertex \((h, k) = (\frac{3}{2}, -\frac{1}{4})\) and \(4a = 1 \implies a = \frac{1}{4}\).
Step 2: Analyze Each Point:
1. Q (Point where tangent is parallel to X-axis): This is the vertex. \(Q = (\frac{3}{2}, -\frac{1}{4})\). Matches II. 2. S (Focus): For vertical parabola \((x-h)^2 = 4a(y-k)\), focus is \((h, k+a)\). \(S = (\frac{3}{2}, -\frac{1}{4} + \frac{1}{4}) = (\frac{3}{2}, 0)\). Matches III. 3. Z (Intersection of axis and directrix): The axis is \(x = \frac{3}{2}\). The directrix is \(y = k - a = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}\). Intersection \(Z = (\frac{3}{2}, -\frac{1}{2})\). Matches IV. 4. P (Endpoint of Latus Rectum): The latus rectum passes through the focus (\(y=0\)). Substitute \(y=0\) into parabola equation: \(0 = x^2 - 3x + 2 \implies (x-1)(x-2) = 0 \implies x = 1, 2\). Points are \((1, 0)\) and \((2, 0)\). \((2, 0)\) is in the list. Matches I.
Step 3: Match:
A(P) \(\to\) I B(Q) \(\to\) II C(S) \(\to\) III D(Z) \(\to\) IV