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