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प्रश्न
Find the equation of the tangent to the ellipse 2x2 + y2 = 6 from the point (2, 1).
उत्तर
Given equation of the ellipse is 2x2 + y2 = 6.
∴ `x^2/3 + y^2/6` = 1
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 3 and b2 = 6
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 (2, 1) lies on both the tangents,
1 = `2"m" ± sqrt(3"m"^2 + 6)`
∴ 1 – 2m = `± sqrt(3"m"^2 + 6)`
Squaring both the sides, we get
1 – 4m + 4m2 = 3m2 + 6
∴ m2 – 4m – 5 = 0
∴ (m – 5)(m + 1) = 0
∴ m = 5 or m = – 1
These are the slopes of the required tangents.
∴ By slope point form y – y1 = m(x – x1), the equations of the tangents are
∴ y – 1 = 5(x – 2) and y – 1 = – 1(x – 2)
∴ y – 1 = 5x – 10 and y – 1 = – x + 2
∴ 5x – y – 9 = 0 and x + y – 3 = 0.
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