Example 1 : <\/span> Find the equation of ellipse whose foci are (2, 3), (-2, 3) and whose semi major axis is of length \\(\\sqrt{5}\\)<\/p>\n Solution : <\/span>Here S = (2, 3) & S’ is (-2, 3) and b = \\(\\sqrt{5}\\) \\(\\implies\\) SS’ = 4 = 2ae \\(\\implies\\) ae = 2 Example 2 : <\/span> For what value of k does the line y = x + k touches the ellipse \\(9x^2 + 16y^2\\) = 144.<\/p>\n Solution : <\/span>\\(\\because\\) Equation of ellipse is \\(9x^2 + 16y^2\\) = 144 or \\(x^2\\over 16\\) + \\({(y-3)}^2\\over 9\\) = 1 Example 3 : <\/span> Find the equation of the tangents to the ellipse \\(3x^2+4y^2\\) = 12 which are perpendicular to the line y + 2x = 4<\/p>\n Solution : <\/span>Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4 Practice these given ellipse examples to test your knowledge on concepts of ellipse.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some ellipse examples for better understanding of ellipse concepts. Example 1 : Find the equation of ellipse whose foci are (2, 3), (-2, 3) and whose semi major axis is of length \\(\\sqrt{5}\\) Solution : Here S = (2, 3) & S’ is (-2, 3) and b = \\(\\sqrt{5}\\) \\(\\implies\\) SS’ …<\/p>\n
\n but \\(b^2\\) = \\(a^2(1-e^2)\\) \\(\\implies\\) 5 = \\(a^2\\) – 4 \\(\\implies\\) a = 3
\n Hence the equation to major axis is y = 3.
\n Centre of ellipse is midpoint of SS’ i.e. (0, 3)
\n \\(\\therefore\\) Equation to ellipse is \\(x^2\\over a^2\\) + \\({(y-3)}^2\\over b^2\\) = 1 or \\(x^2\\over 9\\) + \\({(y-3)}^2\\over 5\\) = 1<\/p>\n
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\n comparing this with \\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1 then we get \\(a^2\\) = 16 and \\(b^2\\) = 9
\n and comparing the line y = x + k with y = mx + c m = 1 and c = k
\n If the line y = x + k touches the ellipse \\(9x^2 + 16y^2\\) = 144, then \\(c^2\\) = \\(a^2m^2 + b^2\\)
\n \\(\\implies\\) \\(k^2\\) = 16 \\(\\times\\) \\(1^2\\) + 9 \\(\\implies\\) \\(k^2\\) = 25
\n \\(\\therefore\\) k = \\(\\pm\\)5\n <\/p>
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\\(\\therefore\\) mx – 2 = -1 \\(\\implies\\) m = \\(1\\over 2\\)
Since \\(3x^2+4y^2\\) = 12 or \\(x^2\\over 4\\) + \\(y^2\\over 3\\) = 1
Comparing this with \\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1
\\(\\therefore\\) \\(a^2\\) = 4 and \\(b^2\\) = 3
So the equation of the tangent are y = \\(1\\over 2\\)x \\(\\pm\\) \\(\\sqrt{4\\times {1\\over 4} + 3}\\)
\\(\\implies\\) y = \\(1\\over 2\\)x \\(\\pm\\) 2 or x – 2y \\(\\pm\\) 4 = 0<\/p>\n
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