JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The locus of the mid-point of the portion intercepted between the axes by the line

x\cos \alpha + y\sin \alpha = p,

where

p

is a constant is

View answer and explanation

Correct option: D

\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = \frac{2}{{{p^2}}}

Option B (

\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = \frac{4}{{{p^2}}}

) is correct because it represents the locus of the midpoint derived from the given line equation.

More Questions from STRAIGHT LINES

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x-3 y=p, a x+2 y=q

a x+y=r

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

(3,8)

x+3 y=7

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0