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प्रश्न
The line `x/a + y/b` = 1 moves in such a way that `1/a^2 + 1/b^2 = 1/c^2`, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2.
पर्याय
True
False
उत्तर
This statement is True.
Explanation:
The given equation is `x/b - y/a` = 0 .......(i)
Equation of the line passing through (0, 0) and perpendicular to equation (i) is
`x/b - y/a` = 0 .....(ii)
Squaring and adding equation (i) and (ii) we get
`(x/a + y/b)^2 + (x/b - y/a)^2` = 1 + 0
⇒ `x^2/a^2 + y^2/b^2 + (2xy)/(ab) + x^2/b^2 + y^2/a^2 - (2xy)/(ab)` = 1
⇒ `x^2(1/a^2 + 1/b^2) + y^2(1/b^2 + 1/a^2)` = 1
⇒ `(x^2 + y^2) (1/a^2 + 1/b^2)` = 1
⇒ `(x^2 + y^2)(1/c^2)` = 1 ....`[because 1/a^2 + 1/b^2 = 1/c^2]`
⇒ x2 + y2 = c2
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