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Find the equation of ellipse, Centre at (0,0), major axis on the y-axis and passes through the points (3,2) and (1,6). |
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Answer» Since the major axis of the ellipse lies on the y-axis, it is a vertical ellipse. Let the required equation be `x^(2)/b^(2) + y^(2)/a^(2) = 1" "("where " a^(2) gt b^(2))." "` …(i) Since (3, 2) lies on (i) , we have ` 9/b^(2) + 4/a^(2) = 1." "` ...(ii) Also, since (1, 6) lies on(i), we have `1/b^(2) + 36/a^(2) = 1." " ` ...(iii) Putting ` 1/b^(2) = u and 1/a^(2) = v`, these equations become: ` 9u+4v = 1 " "` ...(iv) ` and u + 36 v = 1" " ` (v) On multiplying (v) by 0 and subtracting (iv) from it, we get ` 320v = 8 hArr v = 8/(320) = 1/(40) hArr 1/a^(2) = 1/40 hArr a^(2) = 40.` Putting ` v = 1/40 ` in (v), we get ` u + (36 xx 1/40) = 1 hArr u = (1-9/10) = 1/10 hArr 1/b^(2) = 1/10 hArr b^(2) = 10.` Thus, ` b^(2) and a^(2) = 40.` Hence, the required equation is ` x^(2)/10 + y^(2) / 40 = 1.` |
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