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Locus of Intersection of Perpendicular Tangents to x^2 - 8x + 2y + 2 = 0 | JEE Advanced Parabola

JEE Maths question with a full step-by-step solution.

Question
Two perpendicular tangents to the parabola x2−8x+2y+2=0x^{2} - 8x + 2y + 2 = 0 meet at a point. Find the locus of that point.
A2y−15=02y - 15 = 0correct
B2y+15=02y + 15 = 0
C2x+9=02x + 9 = 0
Dnone of these
Solution
Question attachment Step 1: For any parabola the locus of the point of intersection of two perpendicular tangents is its directrix. Step 2: Completing the square,
x2−8x=−2y−2 ⟹ x2−8x+16=−2y−2+16x^{2}-8x = -2y-2 \ \Longrightarrow\ x^{2}-8x+16 = -2y-2+16
(x−4)2=−2(y−7)(x-4)^{2} = -2(y-7)
 Step 3: Comparing with (x−h)2=−4a(y−k)(x-h)^{2} = -4a(y-k),
h=4,k=7,4a=2 ⟹ a=12h=4, \quad k=7, \quad 4a = 2 \ \Longrightarrow\ a = \tfrac12
 so the vertex is (4,7)(4,7) and the parabola opens downwards. Step 4: For a downward-opening parabola the directrix lies at a distance aa above the vertex:
y=7+12=152 ⟹ 2y−15=0y = 7 + \frac12 = \frac{15}{2} \ \Longrightarrow\ 2y - 15 = 0
 Answer: (1).
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