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प्रश्न
Find the equation of the ellipse in standard form if passing through the points (−3, 1) and (2, −2)
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उत्तर
Let the required equation of ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1, where a > b.
The ellipse passes through the points (–3, 1) and (2, –2).
∴ Substituting x = –3 and y = 1 in equation of ellipse, we get
`(-3)^2/"a"^2 + 1^2/"b"^2` = 1
∴ `9/"a"^2 + 1/"b"^2` = 1 ...(i)
Substituting x = 2 and y = –2 in equation of ellipse, we get
`2^2/"a"^2 + (-2)^2/"b"^2` = 1
∴ `4/"a"^2 + 4/"b"^2` = 1 ...(ii)
Let `1/"a"^2` = A and `1/"b"^2` = B
∴ Equations (i) and (ii) become
9A + B = 1 …(iii)
4A + 4B = 1 …(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …(v)
Subtracting (iv) from (v), we get
32A = 3
∴ A = `3/32`
Substituting A = `3/32` in (iv), we get
`4(3/32) + 4"B"` = 1
∴ `3/8 + 4"B"` = 1
∴ 4B = `1 - 3/8`
∴ 4B = `5/8`
∴ B = `5/32`
Since `1/"a"^2` = A and `1/"b"^2` = B,
`1/"a"^2 = 3/32` and `1/"b"^2 = 5/32`
∴ a2 = `32/3` and b2 = `32/5`
∴ The required equation of ellipse is
`x^2/((32/3)) + y^2/((32/5))`, i.e., 3x2 + 5y2 = 32.
संबंधित प्रश्न
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