Question

If the equation of directrix of an ellipse x^2/a^2 + y^2/b^2 = 1 is x = 4, then normal to the ellipse at point (1, B).(B ≥ 0) passes through the point (where eccentricity of the ellipse is 1/2 )

Answer 1

Answer:

2) is your answer.

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Answer 2

ANSWER.

=> The normal of the ellipse at point ( 1 , b)

( b > 0 ) passes through the point = ( 1, 3/2 )

Option [ 1 ] is correct answer.

EXPLANATION.

$\sf \to \: equation \: of \: directrix \: of \: an \: ellipse \\ \\ \sf \to \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \: \: \: is \: \: \: x = 4 \\ \\ \sf \to \: eccentricity \: = \frac{1}{2} \: \: \: (given)$

$\sf \to \: equation \: of \: directrix \: = x \: = \dfrac{a}{e} \\ \\ \sf \to \: 4 = \frac{a}{ \dfrac{1}{2} } \implies \: 4 = 2a \\ \\ \sf \to \: a \: = 2 \implies \: {a}^{2} = 4 \\ \\ \sf \to \: eccentricity \: of \: the \: ellipse \: \implies \: {b}^{2} = {a}^{2} (1 - {e}^{2}) \\ \\ \sf \to \: {b}^{2} = 4(1 - ( \frac{1}{2}) {}^{2} ) \\ \\ \sf \to \: {b}^{2} = 4(1 - \frac{1}{4}) \\ \\ \sf \to \: {b}^{2} = 4( \frac{3}{4}) \implies \: {b}^{2} = 3$

$\sf \to \: equation \: of \: ellipse \: is \: written \: as \\ \\ \sf \to \: \frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{3} = 1$

$\sf \to \: the \: normal \: to \: the \: ellipse \: at \: pont \: (1, \beta ) \: \: ( \beta > 0) \\ \\ \sf \to \: put \: the \: value \: of \: normal \: in \: the \: equation \: of \: ellipse \\ \\ \sf \to \: \frac{ {1}^{2} }{4} + \frac{ { \beta }^{2} }{3} = 1 \\ \\ \sf \to \: \frac{ { \beta }^{2} }{3} = 1 - \frac{ 1 }{4} \\ \\ \sf \to \: \frac{ { \beta }^{2} }{3} = \frac{3}{4} \\ \\ \sf \to \: \beta = \frac{3}{2} \: \: and \: \: ( \beta > 0)$

$\sf \to \: \green{{ \underline{normal \: at \: the \: ellipse \: passes \: through \: point \: = (1, \dfrac{3}{2} }}})$