Ex 11.3, 19 - Class 11

Transcript

Ex 11.3, 19 Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6). Since major axis is along y-axis & centre is at (0,0) So required equation of ellipse is ﷐﷐𝒙﷮𝟐﷯﷮﷐𝒃﷮𝟐﷯﷯ + ﷐﷐𝒚﷮𝟐﷯﷮﷐𝒂﷮𝟐﷯﷯ = 1 Given that ellipse passes through point (3, 2) & (1, 6) Points (3, 2) & (1, 6) will satisfy equation of ellipse. Putting x = 3 & y = 2 in (1) ﷐﷐﷐3﷯﷮2﷯﷮﷐𝑏﷮2﷯﷯ + ﷐﷐﷐2﷯﷮2﷯﷮﷐𝑎﷮2﷯﷯ = 1 ﷐9﷮﷐𝑏﷮2﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 Putting x = 1 & y = 6 in (1) ﷐﷐(1)﷮2﷯﷮﷐𝑏﷮2﷯﷯ + ﷐﷐(6)﷮2﷯﷮﷐𝑎﷮2﷯﷯ = 1 ﷐1﷮﷐𝑏﷮2﷯﷯ + ﷐36﷮﷐𝑎﷮2﷯﷯ = 1 Now equations are ﷐9﷮﷐𝑏﷮2﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 …(2) ﷐1﷮﷐𝑏﷮2﷯﷯ + ﷐36﷮﷐𝑎﷮2﷯﷯ = 1 …(3) From (3) ﷐1﷮﷐𝑏﷮2﷯﷯ + ﷐36﷮﷐𝑎﷮2﷯﷯ = 1 ﷐1﷮﷐𝑏﷮2﷯﷯ = 1 − ﷐36﷮﷐𝑎﷮2﷯﷯ Putting value of b2 in (2) ﷐9﷮﷐𝑏﷮2﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 9﷐﷐1﷮﷐𝑏﷮2﷯﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 9﷐1−﷐36﷮﷐𝑎﷮2﷯﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 9 − ﷐324﷮﷐𝑎﷮2﷯﷯ + ﷐4﷮﷐𝑎﷮2﷯﷯ = 1 ﷐−320﷮﷐𝑎﷮2﷯﷯ = 1 − 9 ﷐−320﷮﷐𝑎﷮2﷯﷯ = −8 ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐−8﷮−320﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐8﷮320﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐1﷮40﷯ So, a2 = 40 Putting value of ﷐𝑎﷮2﷯ in (3) ﷐1﷮﷐𝑏﷮2﷯﷯ + ﷐36﷮﷐𝑎﷮2﷯﷯ = 1 ﷐1﷮﷐𝑏﷮2﷯﷯ = 1 − ﷐36﷮﷐𝑎﷮2﷯﷯ ﷐1﷮﷐𝑏﷮2﷯﷯ = 1 − 36 ﷐﷐1﷮40﷯﷯ ﷐1﷮﷐𝑏﷮2﷯﷯ = ﷐40 −36﷮40﷯ ﷐1﷮﷐𝑏﷮2﷯﷯ = ﷐4﷮40﷯ ﷐1﷮﷐𝑏﷮2﷯﷯ = ﷐1﷮10﷯ So, b2 = 10 Now required equation of ellipse is ﷐﷐𝑥﷮2﷯﷮﷐𝑏﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑎﷮2﷯﷯ = 1 Putting value of b2 & a2 ﷐﷐𝒙﷮𝟐﷯﷮𝟏𝟎﷯ + ﷐﷐𝒚﷮𝟐﷯﷮𝟒𝟎﷯ = 1