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BITSAT 2014 Paper

Section: Mathematics

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Question : 124 of 150

Marks: +1, -0

Through the vertex O of a parabola

y^2

= 4x, chords OP and OQ are drawn at right angles to one another. The locus of the middle point of PQ is

$y^2$ = 2x + 8
$y^2$ = x + 8
$y^2$ = 2x – 8
$y^2$ = x – 8

Solution:

Given parabola is

y^2

= 4x ...(1)

Lt P =

(t_1^2, 2t_1)

and Q =

(t_2^2, 2t_2)

Slope of OP =

\frac{2t_1}{t_1^2}

=

\frac{2}{t_1}

and slope of OQ =

\frac{2}{t_2}

Since OP ⊥ OQ,

∴

\frac{4}{t_1 t_2}

= - 1 or

t_1 t_2

= - 4 ... (2)

Let R (h, k) be the middle point of PQ, then

h =

\frac{t_1^2 + t_2^2}{2}

... (3)

and k =

t_1 + t_2

... (4)

From (4),

k^2

=

t_1^2 + t_2^2 + t_1 t_2

= 2h - 8

[ From (2) and (3)]

Hence locus of R (h, k) is

y^2

= 2x – 8.

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