AP EAMCET · Maths · Parabola
If the perpendicular distance from the focus of a parabola \(\mathrm{y}^2=4 \mathrm{ax}\) to its directrix is \(\frac{3}{2}\), then the equation of the normal drawn at \((4 a,-4 a)\) is
- A \(2 x+y=3\)
- B \(2 x-y=9\)
- C \(x-2 y=9\)
- D \(x+2 y+3=0\)
Answer & Solution
Correct Answer
(B) \(2 x-y=9\)
Step-by-step Solution
Detailed explanation
\(2a = \frac{3}{2} \implies a = \frac{3}{4}\) From \((at^2, 2at) = (4a, -4a)\), we get \(t = -2\). Normal: \(y = -tx + 2at + at^3\) \(y = -(-2)x + 2(\frac{3}{4})(-2) + (\frac{3}{4})(-2)^3\) \(y = 2x - 3 - 6\) \(2x - y = 9\)
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