If the angular bisectors of the coordinate axes cut the parabola $y^{2} = 4 a x$ at the points $O, A, B$ then the area of $Δ O A B$ is ( $O$ is the origin)

32 a^{2}

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16 a^{2}

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64 a^{2}

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8 a^{2}

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Solution

The correct option is A $16 a^{2}$
Angular bisectors of coordinate axes are, $y = x$ and $y = - x$
And given parabola is $y^{2} = 4 a x$
Now solving above three equation we get, $A = (4 a, 4 a), B = (4 a, - 4 a), O = (0, 0)$
Hence are of triangle OAB is $= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 4 a & 4 a & 1 4 a & - 4 a & 1 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 16 a^{2}$

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Similar questions

A, B, C

y^{2} = 4 a x

G . P

A

C

A = (a C o s θ, b S i n θ)

B = (- a S i n θ, b C o s θ)

O

θ

Δ A

B

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =

3 x + 4 y - 24 = 0

Δ O A B

y^{2} = 4 a x, x^{2} = 4 a y

y^{2} = 4 a x

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