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Prove that the product of the lengths of the perpendiculars drawn from the points `(sqrt(a^2-b^2),0)`and `(-sqrt(a^2-b^2),0)`to the line `x/a``costheta``+``y/b``sintheta=1`is `b^2`.

Text Solution

Verified by Experts

The equation of the given line is
`(x)/(a)"cos" theta + (y)/(b) "sin" theta = 1`
`" or " bx "cos" theta + ay " sin " theta-ab = 0" " (1)`
`" The length of the perpendicular from the point" (sqrt(a^(2)-b^(2)),0) "to line" (1) " is "`
`p_(1) = (|b "cos" theta sqrt(a^(2)-b^(2)) + a "sin" theta (0)-ab|)/(sqrt(b^(2) "cos"^(2) theta +a^(2) "sin"^(2) theta))`
`=(|b "cos" theta sqrt(a^(2) -b^(2))-ab|)/(sqrt(b^(2) cos^(2) theta +a ^(2) "sin"^(2) theta)) " " (2)`
`"The length of the perpendicular from the point" (-sqrt(a^(2)-b^(2)),0) " to line (I) is "`
`p_(2) = (|b "cos" theta (-sqrt(a^(2)-b^(2))) + a "sin" theta (0)-ab|)/(sqrt(b^(2) "cos"^(2) theta +a^(2) "sin"^(2) theta))`
`= (|b"cos" theta sqrt(a^(2)-b^(2)) +ab|)/(sqrt(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta)) " " (3)`
`therefore p_(1) p_(2) = (|b"cos" theta sqrt(a^(2)-b^(2)) -ab||b "cos" theta sqrt(a^(2)-b^(2)) +ab|)/(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta)`
`=(|b^(2)"cos"^(2) theta (a^(2)-b^(2)) -a^(2)b^(2)|)/((b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta))`
`=(|a^(2)b^(2)"cos"^(2) theta-b^(4) "cos"^(2)theta-a^(2)b^(2)|)/(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta)`
`=(b^(2)|a^(2)"cos"^(2) theta-b^(2) "cos"^(2)theta-a^(2)|)/(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta)`
`=(b^(2)|-b^(2)"cos"^(2) theta-a^(2)(1- "cos"^(2)theta)|)/(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta)`
`=(b^(2)(b^(2)"cos"^(2) theta + a^(2)"sin"^(2)theta))/(b^(2) "cos"^(2) theta + a^(2) "sin"^(2) theta) =b^(2)`
Hence, proved.
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