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प्रश्न
Find the equation of the tangent to the ellipse x2 + 4y2 = 9 which are parallel to the line 2x + 3y – 5 = 0.
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उत्तर
We know that the equations of tangents with slope m to the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1 are
y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)` ...(1)
The equation of the ellipse is x2 + 4y2 = 9
∴ `x^2/9 + y^2/((9/4)` = 1
Comparing this with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 9, b2 = `9/4`
Slope of 2x + 3y – 5 = 0 is `-2/3`
The required tangent is parallel to it
∴ its slope = m = `-2/3`
Using (1), the required equations of tangents are
y = `-(2x)/3 ± sqrt(9 xx 4/9 + 9/4)`
∴ y = `-(2x)/3 ± sqrt(25/4)`
∴ y = `-(2x)/3 ± 5/2`
∴ 6y = – 4x ± 15
∴ 4x + 6y = ± 15
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संबंधित प्रश्न
Find the
- lengths of the principal axes.
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- length of the latus rectum
- distance between focii
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- distance between focii
- distance between directrices of the ellipse:
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