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प्रश्न
Find the equation of the tangent to the ellipse 4x2 + 7y2 = 28 from the point (3, –2).
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उत्तर
Given equation of the ellipse is 4x2 + 7y2 = 28
∴ `x^2/7 + y^2/4` = 1
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 7 and b2 = 4
Equations of tangents to the ellipse
`x^2/"a"^2 + y^2/"b"^2` = 1 having slope m are
y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`
Since (3, –2) lies on both the tangents,
–2 = `3"m" ± sqrt(7"m"^2 + 4)`
∴ –2 – 3m = `± sqrt(7"m"^2 + 4)`
Squaring both the sides, we get
9m2 + 12m + 4 = 7m2 + 4
∴ 2m2 + 12m = 0
∴ 2m(m + 6) = 0
∴ m = 0 or m = – 6
∴ These are the slopes of the required tangents.
∴ By slope point form y – y1 = m(x – x1), the equations of the tangents are
y + 2 = 0(x – 3) and y + 2 = –6(x – 3)
∴ y + 2 = 0 and y + 2 = –6x + 18
∴ y + 2 = 0 and 6x + y – 16 = 0.
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