मराठी

The line xa+yb = 1 moves in such a way that 1a2+1b2=1c2, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2. - Mathematics

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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

MCQ
चूक किंवा बरोबर
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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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पाठ 10: Straight Lines - Exercise [पृष्ठ १८४]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 10 Straight Lines
Exercise | Q 54 | पृष्ठ १८४

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