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Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse x225+y216 = 1 is equal to 16 - Mathematics and Statistics

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Question

Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse `x^2/25 + y^2/16` = 1 is equal to 16

Sum
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Solution

Given equation of the ellipse is `x^2/25 + y^2/16` = 1.

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

∴ a2 = 25, b2 = 16

∴ a = 5, b = 4

We know that e = `sqrt("a"^2 - "b"^2)/"a"`

∴ e = `sqrt(25 - 16)/5`

= `sqrt(9)/5`

= `3/5`

ae = `5(3/5)`

= 3

Co-ordinates of foci are S(ae, 0) and S'(– ae, 0),

i.e., S(3, 0) and S'(–3, 0)

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)`

Equation of one of the tangents to the ellipse is

y = `"m"x + sqrt(25"m"^2 + 16)`

∴ `"m"x - y + sqrt(25"m"^2 + 16)` = 0   ...(i)

p1 = length of perpendicular segment from S(3, 0) to the tangent (i)

= `|("m"(3) - 0 + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)|`

∴ p1 = `|(3"m" + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)|`

p2 = length of perpendicular segment from S'(–3, 0) to the tangent (i)

= `|("m"(-3) - 0 + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)|`

∴ p2 = `|(-3"m" + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)|`

∴ p1p2 = `|(3"m" + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)| |(-3"m" + sqrt(25"m"^2 + 16))/sqrt("m"^2 + 1)|`

= `((25"m"^2 + 16) - 9"m"^2)/("m"^2 + 1)`

= `(16("m"^2 + 1))/("m"^2 + 1)`

= 16

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Chapter 7: Conic Sections - Exercise 7.2 [Page 163]

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Answer the following:

Find the

  1. lengths of the principal axes
  2. co-ordinates of the foci
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  4. length of the latus rectum
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`x^2/25 + y^2/9` = 1


Find the

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