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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If \(y\, = mx + c\) is the normal at a point on the parabola \(y^2\, = 8x\) whose focal distance is \(8\, units\), then \(\left| c \right|\) is equal to
- A \(2\sqrt 3 \)
- B \(8\sqrt 3 \)
- C \(10\sqrt 3 \)
- D \(16\sqrt 3 \)
Answer & Solution
Correct Answer
(C) \(10\sqrt 3 \)
Step-by-step Solution
Detailed explanation
\(c = - 29m - 9{m^3}\) \(a = 2\) Given \({\left( {a{t^2} + a} \right)^2} + 4{a^2}{t^2} = 64\) \( \Rightarrow \left( {a\left( {{t^2} + 1} \right)} \right) = 8\) \( \Rightarrow {t^2} + 1 = 4\,\, \Rightarrow {t^2} = 3\) \( \Rightarrow t = \sqrt 3 \)…
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