JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The eccentricity of an ellipse whose centre is at the origin is \(\frac{1}{2}\) . If one of its directices is \(x = - 4\) then the equation of the normal to it at \(\left( {1,\frac{3}{2}} \right)\) is
- A \(x + 2y = 4\)
- B \(2y - x = 2\)
- C \(4x - 2y = 1\)
- D \(4x + 2y = 7\)
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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