A tangent and a normal are drawn at the point p(2, –4) on the parabola y2 = 8x, which meet the directrix of the parabola at the point A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :

Engineering

Mathematics

Normal to Parabola

Question

A tangent and a normal are drawn at the point p(2, –4) on the parabola y² = 8x, which meet the directrix of the parabola at the point A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :

– 16

– 12

– 20

– 18

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Equation of tangent at (2, –4) (T = 0)

–4y = 4(x + 2)

x + y + 2 = 0 ….(i)

equation of normal

x – y + λ = 0

$↓$ (2, –4)

λ = –6

thus x – y = 6 ….. (2) equation of normal

POI of (1) & x = –2 is A(–2, 0)

POI of (2) & x = –2 is A(–2, 8)

Given AQBP is a sq.

⇒ m_AQ. m_AP = –1

$\Rightarrow (\frac{b}{a + 2}) (\frac{4}{- 4}) = - 1 \Rightarrow a + 2 = b$ ….(1)

Also PQ must be parallel to x-axis thus

⇒ b = – 4

$∴$ a = –6

Thus 2a + b = – 16

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