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प्रश्न
Find the equation of the tangent to the ellipse `x^2/5 + y^2/4` = 1 passing through the point (2, –2)
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उत्तर
Given equation of the ellipse is `x^2/5 + y^2/4` = 1
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 5 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 (2, – 2) lies on both the tangents,
– 2 = `2"m" ± sqrt(5"m"^2 + 4)`
∴ – 2 – 2m = `± sqrt(5"m"^2 + 4)`
Squaring both the sides, we get
4m2 + 8m + 4 = 5m2 + 4
∴ m2 – 8m = 0
∴ m(m – 8) = 0
∴ m = 0 or m = 8
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 – 2) and y + 2 = 8(x – 2)
∴ y + 2 = 0 and y + 2 = 8x – 16
∴ y + 2 = 0 and 8x – y – 18 = 0
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- co-ordinates of the foci
- equations of directrices
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