- Home
- Standard 11
- Mathematics
9.Straight Line
normal
The line $\frac{x}{a} + \frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -
A
$x^2 + y^2 = c^2$
B
$x^2 + y^2 = 2c^2$
C
$x^2 + y^2 = \frac{c^2}{2}$
D
$x^2 + y^2 = 4c^2$
Solution
${\rm{OP}} = \left| {\frac{1}{{\sqrt {\frac{1}{{{{\rm{a}}^2}}} + \frac{1}{{{{\rm{b}}^2}}}} }}} \right|$
$\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=2 \mathrm{c}^{2}$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=2 \mathrm{c}^{2}$
Standard 11
Mathematics
