Test Index

Parabola Part 2

Section: Mathematics

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Question : 11 of 51

Marks: +1, -0

If the chord joining the points

P _1 (x_1, y_1)

P _2 (x_2, y_2)

on the parabola

y^2=12 x

subtends a right angle at the vertex of the parabola, then

x_1 x_2-y_1 y_2

[22 Jan 2026 Shift 1]

280
288
292
284

Solution:

Concept:

For a parabola

y^2 = 4ax

, the chord joining points

(a t_1^2, 2a t_1)

and

(a t_2^2, 2a t_2)

subtends a right angle at the vertex if

t_1 t_2 = -4

.

Explanation:

The given parabola is

y^2 = 12x

, so

4a = 12

and

a = 3

.

Parametric points:

P_1(3t_1^2, 6t_1)

and

P_2(3t_2^2, 6t_2)

.

Slope of

OP_1

is

\frac{6t_1}{3t_1^2} = \frac{2}{t_1}

, slope of

OP_2

is

\frac{2}{t_2}

.

Since

\angle P_1 O P_2 = 90^\circ

, slopes satisfy

(\frac{2}{t_1})(\frac{2}{t_2}) = -1

.

Thus

\frac{4}{t_1 t_2} = -1

gives

t_1 t_2 = -4

.

Now

x_1 x_2 - y_1 y_2 = (3t_1^2)(3t_2^2) - (6t_1)(6t_2) = 9(t_1 t_2)^2 - 36(t_1 t_2)

.

Substitute

t_1 t_2 = -4

:

9(16) - 36(-4) = 144 + 144 = 288

.

Answer:

288

, which corresponds to option B.

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