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The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity 12 is ______. - Mathematics

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Question

The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity `1/2` is ______.

Options

  • 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0

  • 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0

  • 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0

  • None

MCQ
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Solution

The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity `1/2` is 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0.

Explanation:

Given that focus of the ellipse is (1, – 1)

And the equation of the directrix is x – y – 3 = 0

And e = `1/2`.

Let P(x, y) by any point on the parabola

∴ `"PF"/("Distance of the point P from the directrix")` = e

= `sqrt((x - 1)^2 + (y + 1)^2)/|(x - y - 3)/sqrt((1)^2 + (-1)^2)| = 1/2`

⇒ `2sqrt(x^2 + 1 - 2x + y^2 + 1 + 2y) = |(x - y - 3)/sqrt(2)|`

Squaring both sides, we have

⇒ `4(x^2 + y^2 - 2x + 2y + 2) = (x^2 + y^2 + 9 - 2xy - 6y - 6x)/2`

⇒ 8x2 + 8y2 – 16x + 16y + 16 = x2 + y2 – 2xy + 6y – 6x + 9

⇒ 7x2 + 7y2 + 2xy – 10x + 10y + 7 = 0

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Chapter 11: Conic Sections - Exercise [Page 206]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 11 Conic Sections
Exercise | Q 54 | Page 206

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