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प्रश्न
Find the equation of the tangent to the ellipse 5x2 + 9y2 = 45 which are ⊥ to the line 3x + 2y + y = 0.
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उत्तर
The equations of the tangents to the ellipse
`x^2/"a"^2 + y^2/"b"^2` = 1 in terms of slope m are
y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)` ...(1)
The equation of the ellipse is
5x2 + 9y2 = 45,
i.e., `x^2/9 + y^2/5` = 1
Comparing this with `x^2/"a"^2 + y^2/"b"^2` = 1, we get,
a2 = 9, b2 = 5
Slope of the line 3x + 2y + 7 = 0 is `-3/2`.
Since the tangent is perpendicular to this line,
its slope = m = `2/3`
Using (1), the required equations of tangents are
y = `2/3x ± sqrt(9 xx 4/9 + 5)`
∴ y = `2/3x ± 3`
∴ 3y = 2x ± 9
∴ 3y – 2x = ± 9.
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संबंधित प्रश्न
Answer the following:
Find the
- lengths of the principal axes
- co-ordinates of the foci
- equations of directrices
- length of the latus rectum
- distance between foci
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Find the
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- co-ordinates of the focii
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- length of the latus rectum
- distance between focii
- distance between directrices of the ellipse:
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