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STD 11 - 9. straight line — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9. straight line4 Marks
MCQ
If for a variable line $\frac{x}{a} + \frac{y}{b} = 1$, the condition $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}$ (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is
  • A
    ${x^2} + {y^2} = {c^2}/2$
  • B
    ${x^2} + {y^2} = 2{c^2}$
  • ✓
    ${x^2} + {y^2} = {c^2}$
  • D
    ${x^2} - {y^2} = {c^2}$
✓

Answer

Correct option: C.
${x^2} + {y^2} = {c^2}$
c
(c) Equation of perpendicular drawn from origin to the line $\frac{x}{a} + \frac{y}{b} = 1$ is $y - 0 = \frac{a}{b}(x - 0)$

[ $m$ of given line $ = \frac{{ - b}}{a}$, $\therefore m$ of perpendicular $\left. { = \frac{a}{b}} \right]$

==> $by - ax = 0$

==> $\frac{x}{b} - \frac{y}{a} = 0$

Now, the locus of foot of perpendicular is the intersection point of line $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$

and $\frac{x}{b} - \frac{y}{a} = 0$......$(ii)$

To find locus, squaring and adding $(i)$ and $(ii)$

${\left( {\frac{x}{a} + \frac{y}{b}} \right)^2} + {\left( {\frac{x}{b} - \frac{y}{a}} \right)^2} = 1$

==> ${x^2}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + {y^2}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) = 1$

==> ${x^2}\left( {\frac{1}{{{c^2}}}} \right) + {y^2}\left( {\frac{1}{{{c^2}}}} \right) = 1$ ,

==> ${x^2} + {y^2} = {c^2}$.


 

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