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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

एक दीर्घवृत्त, जिसका केन्द्र मूल बिन्दु पर है, की उत्केन्द्रता \(\frac{1}{2}\) है। यदि उसकी एक नियता \(x=-4\) है, तो उसके बिंदु \(\left(1, \frac{3}{2}\right)\) पर उसके अभिलंब का समीकरण है:

  1. A \(x + 2y = 4\)
  2. B \(2y - x = 2\)
  3. C \(4x - 2y = 1\)
  4. D \(4x + 2y = 7\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4x - 2y = 1\)

Step-by-step Solution

Detailed explanation

Eccentricity of ellipse \( = \frac{1}{2}\) Now, \( - \frac{a}{e} = - 4 \Rightarrow a = 4 \times \frac{1}{2} = 2 \Rightarrow a = 2\) we have \({b^2} = {a^2}\left( {1 - {e^2}} \right) = {a^2}\left( {1 - \frac{1}{4}} \right) = 4 \times \frac{3}{4} = 3\) \(\therefore \) Equation of…
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