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NDA 2020 Maths · Mathematics · Conic Sections · easy

The directrix of the parabola x^2+6x-8y+25=0 is the line

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The directrix of the parabola x2+6x−8y+25=0x^2+6x-8y+25=0 is the line
  1. A. y=4y=4
  2. B. y=0y=0
  3. C. x=−1x=-1
  4. D. x=−5x=-5
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Answer: (B) y=0y=0

Complete the square: (x+3)2=8y−16=8(y−2)(x+3)^2=8y-16=8(y-2). This is X2=4aYX^2=4aY with 4a=84a=8, so a=2a=2, vertex (−3,2)(-3,2), and the parabola opens upwards. The directrix lies aa below the vertex, so it is y=2−2=0y=2-2=0. The line y=4y=4 is the latus-rectum line through the focus (−3,4)(-3,4), and the vertical lines x=−1x=-1 and x=−5x=-5 are parallel to the axis, which cannot be directrices of this upward parabola.
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