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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
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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
RELATED QUESTIONS
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