Chapter 11 Class 11 Conic Sections

Class 11
Important Questions for exams Class 11

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### Transcript

Ex 10.3, 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 π^π/π^π + π^π/π^π = 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/π^2 + (2)^2/π^2 = 1 9/π^2 + 4/π^2 = 1 Putting x = 1 & y = 6 in (1) γ(1)γ^2/π^2 + γ(6)γ^2/π^2 = 1 1/π^2 + 36/π^2 = 1 From (3) 1/π^2 + 36/π^2 = 1 1/π^2 = 1 β 36/π^2 Putting value of b2 in (2) 9/π^2 + 4/π^2 = 1 9(1/π^2 ) + 4/π^2 = 1 9(1β36/π^2 ) + 4/π^2 = 1 9 β 324/π^2 + 4/π^2 = 1 (β320)/π^2 = 1 β 9 (β320)/π^2 = β8 1/π^2 = (β8)/(β320) 1/π^2 = 8/320 1/π^2 = 1/40 a2 = 40 Putting value of π^2 in (3) 1/π^2 + 36/π^2 = 1 1/π^2 = 1 β 36/π^2 1/π^2 = 1 β 36 (1/40) 1/π^2 = (40 β 36)/40 1/π^2 = 4/40 1/π^2 = 1/10 b2 = 10 Now required equation of ellipse is π₯^2/π^2 + π¦^2/π^2 = 1 Putting value of b2 & a2 π^π/ππ + π^π/ππ = 1