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प्रश्न
Prove that the product of the lengths of the perpendiculars drawn from the points `(sqrt(a^2 - b^2), 0)` and `(-sqrta^2-b^2, 0)` to the line `x/a cos theta + y/b sin theta = 1` is `b^2`.
उत्तर
Given line `"x"/"a" cos θ + "y"/"b" sin θ - 1 = 0` .......(i)
Length of perpendicular drawn from `(sqrt("a"^2 - "b"^2), 0)`
p1 = `(sqrt("a"^2 - "b"^2)/"a" cos θ - 1)/ sqrt(cos^2 θ/"a"^2 + (sin^2 θ)/"b"^2)` .............`[∵ "d" = ("ax"_1 + "by"_1 + "c"_1)/sqrt("a"^2 + "b"^2)]`
Similarly, the length of the perpendicular drawn from `(- sqrt("a"^2 - "b"^2), 0)` to line (i)
p2 = `(-sqrt("a"^2 - "b"^2)/"a" cos θ - 1)/ sqrt(cos^2 θ/"a"^2 + (sin^2 θ)/"b"^2)`
∴ P1P2 = `(sqrt("a"^2 - "b"^2)/"a" cos θ - 1)/ sqrt(cos^2 θ/"a"^2 + (sin^2 θ)/"b"^2) xx (-sqrt("a"^2 - "b"^2)/"a" cos θ - 1)/ sqrt(cos^2 θ/"a"^2 + (sin^2 θ)/"b"^2)`
= `(-(("a"^2 - "b"^2)/"a"^2 cos^2 θ - 1))/(cos ^2 θ/"a"^2 + (sin^2 θ)/"b"^2)`
= `(-"b"^2 [("a"^2 - "b"^2) cos^2 θ - "a"^2])/("b"^2 cos^2 θ + "a"^2 sin^2 θ)`
= `(-"b"^2 [("a"^2 - "b"^2) cos^2 θ - "a"^2])/("b"^2 cos^2 θ+ "a"^2 (1 - cos^2 θ))` ..........[∴ sin2 θ = 1 - cos2 θ]
= `(-"b"^2 [("a"^2 - "b"^2) cos^2 θ - "a"^2])/(("b"^2 - "a"^2) cos^2 θ + "a"^2)`
= `(-"b"^2 [("a"^2 - "b"^2) cos^2 θ - "a"^2])/(-("a"^2 - "b"^2) cos^2 θ - "a"^2)`
= `"b"^2`
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