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Parabola

Section: Mathematics

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Question : 3 of 15

Marks: +1, -0

The lengths of the two focal chords of the parabola

y^2=16 x

is 25 units each. If these two chords cut the parabola at A, B, C and D, then the area (in sq. units) of the quadrilateral formed by

A, B, C

[AP EAPCET 23 May 2025 S1]

$\;\frac{625}{2}$
$180$
$150$
$300$

Solution:

Given equation of parabola is

y^2=16 x

Comparing it with

y^2=4 a x

we get,

a=4

Let coordinates of end points of focal chord of a parabola (i) is,

(a t^2, 2 a t)

and

\left(\;\frac{a}{t^2},\;\frac{-2 a}{t}\right)

\;=\;\text{ its length }\;=\sqrt{ \left(a t^2-\frac{a}{t^2}\right)^2 + \left(2 a t+\frac{2 a}{t}\right)^2 }\;\text{ (given) }\;

\;=25

\;\Rightarrow \sqrt{ \left(4 \left(t^2-\frac{1}{t^2}\right)\right)^2 + \left(8 \left(t+\frac{1}{t}\right)\right)^2 } = 25

(putting

a=4

)

\;\Rightarrow 16 \left(t^2-\frac{1}{t^2}\right)^2 + 64 \left(t+\frac{1}{t}\right)^2 = 625

\;\Rightarrow 16 \left(t^4+\frac{1}{t^4}-2\right) + 64 \left(t^2+\frac{1}{t^2}+2\right) = 625

Let

t+\frac{1}{t}=u

so that

\left(t+\frac{1}{t}\right)^2 = u^2

\;\Rightarrow \;\; t^2+\frac{1}{t^2}+2 = u^2

\;\Rightarrow \;\; t^2+\frac{1}{t^2} = u^2-2

\;\Rightarrow \;\; \left(t^2+\frac{1}{t^2}\right)^2 = \left(u^2-2\right)^2

\;\Rightarrow \;\; t^4+\frac{1}{t^4}+2 = u^4+4-2 u^2

Putting these values in Eq. (ii) and solving then we get,

t=2, t=\frac{1}{2}, t=-2, t=\frac{-1}{2}

So, Points are

A(16,16), B(1,4), C(1,-4), D(16,-16)

∴ Area of quadrilateral

A B C D=

\;\;\frac{1}{2} \left| x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_1) \right|

\;=\;\frac{1}{2} \left| 16 \times 4 + 1 \times (-4) + 1 \times (-16) + 16 \times 16 - (16 \times 1 + 4 \times 1 + (-4) \times 16 + (-16)(16) \right|

\;=300 \;\text{ sq. units }\;

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