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Find the equation of the tangent to the ellipse x2 + 4y2 = 20, ⊥ to the line 4x + 3y = 7.

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प्रश्न

Find the equation of the tangent to the ellipse x2 + 4y2 = 20, ⊥ to the line 4x + 3y = 7.

बेरीज
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उत्तर

Given equation of the ellipse is x2 + 4y2 = 20.

∴ `x^2/20 + y^2/5` = 1

Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get

a2 = 20 and b2 = 5

Slope of the given line 4x + 3y = 7 is `(-4)/3`.

Since the given line is perpendicular to the required tangents, slope of the required tangents is m = `3/4`.

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)`

∴ y = `3/4x ± sqrt(20(3/4)^2 + 5)`

∴ y = `3/4x ± sqrt(45/4 + 5)`

∴ y = `3/4x ± sqrt(65)/2`

∴ 4y = `3x ± 2sqrt(65)`

∴ 3x – 4y = `±2sqrt(65)`

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पाठ 7: Conic Sections - Exercise 7.2 [पृष्ठ १६३]

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