The area of the triangle formed by the tangent and the normal to the parabola $y^{2} = 4 a x$ , both drawn at the same end of the latus rectum, and the axis of the parabola is

2 \sqrt{2} a^{2}

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2 a^{2}

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4 a^{2}

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4 \sqrt{2} a^{2}

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Solution

The correct option is D $4 a^{2}$
$2 a \times y = 2 a (x + a)$
$y = x + a$ (tangent)
$(y - 2 a) = - 1 (x - a)$
$y + x = 3 a$ (Normal)
Aera of $Δ A B C = \frac{1}{2} 4 a \times 2 a$
$= 4 a^{2}$

Suggest Corrections

Similar questions

(x - 2 a)^{2} + y^{2} = x^{2}

y^{2} = 8 x

Δ_{1}

P (\frac{1}{2}, 2)

Δ_{2}

P

\frac{Δ_{1}}{Δ_{2}}

y^{2} = 4 a (x - a) .

y^{2} = 4 a x,

y^{2} = 4 a x

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