Ex 11.3, 20 - Find ellipse: Major axis on x-axis, passes - Ellipse - Defination

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise

Transcript

Ex 11.3, 20 Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2). Since Major axis is on the x-axis So required equation of ellipse is ﷐﷐𝒙﷮𝟐﷯﷮﷐𝒂﷮𝟐﷯﷯ + ﷐﷐𝒚﷮𝟐﷯﷮﷐𝒃﷮𝟐﷯﷯ = 1 Given that ellipse passes through point (4, 3) & (6, 2) Points (4, 3) & (6, 2) will satisfy the equation of ellipse Putting x = 4 & y = 3 in (1) ﷐﷐(4)﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐(3)﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐16﷮﷐𝑎﷮2﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 Putting x = 6 & y = 2 in (1) ﷐﷐(6)﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐(2)﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐36﷮﷐𝑎﷮2﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 Now equation are ﷐16﷮﷐𝑎﷮2﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 …(2) ﷐36﷮﷐𝑎﷮2﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 …(3) From (3) ﷐16﷮﷐𝑎﷮2﷯﷯ = 1 − ﷐9﷮﷐𝑏﷮2﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮16﷯ ﷐1 − ﷐9﷮﷐𝑏﷮2﷯﷯﷯ Putting value of ﷐1﷮﷐𝑎﷮2﷯﷯ in (2) ﷐36﷮﷐𝑎﷮2﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 36﷐﷐1﷮﷐𝑎﷮2﷯﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 36﷐﷐1﷮16﷯﷐1−﷐9﷮﷐𝑏﷮2﷯﷯﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 ﷐36﷮16﷯﷐1−﷐9﷮﷐𝑏﷮2﷯﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 ﷐9﷮4﷯﷐1−﷐9﷮﷐𝑏﷮2﷯﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 ﷐9﷮4﷯ − ﷐81﷮﷐𝑏﷮2﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 ﷐−81﷮4﷐𝑏﷮2﷯﷯ + ﷐4﷮﷐𝑏﷮2﷯﷯ = 1 − ﷐9﷮4﷯ ﷐−81 + 16﷮4﷐𝑏﷮2﷯﷯ = ﷐4 − 9﷮4﷯ ﷐−65﷮4﷐𝑏﷮2﷯﷯ = ﷐−5﷮4﷯ ﷐−5﷮4﷯ ﷐﷐13﷮﷐𝑏﷮2﷯﷯﷯= ﷐−5﷮4﷯ ﷐13﷮﷐𝑏﷮2﷯﷯ = 1 ﷐1﷮﷐𝑏﷮2﷯﷯ = ﷐1﷮13﷯ b2 = 13 Putting value of b2 in ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮16﷯﷐1 − ﷐9﷮﷐𝑏﷮2﷯﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮16﷯﷐1 − ﷐9﷮13﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮16﷯﷐ ﷐13 − 9﷮13﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮16﷯﷐ ﷐4﷮13﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮4 × 16﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮52﷯ a2 = 52 Equation of ellipse is ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 Putting value of a2 & b2 ﷐﷐𝒙﷮𝟐﷯﷮𝟓𝟐﷯ + ﷐﷐𝒚﷮𝟐﷯﷮𝟏𝟑﷯ = 1

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