Consider the parabola y2=8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point P(21,2) on the parabola and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of latus rectum. Δ2Δ1 is :

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Consider the parabola {y}^{{2}}={8}{x}. Let _{{1}} be the area

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Consider the parabola $y^{2} = 8 x .$ Let $Δ_{1}$ be the area of the triangle formed by the end points of its latus rectum and the point P( $\frac{1}{2}$ ,2) on the parabola and $Δ_{2}$ be the area of the triangle formed by drawing tangents at P and at the end points of latus rectum. $\frac{Δ _{1}}{Δ _{2}}$ is :

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Jack Discussed

Consider the parabola

y^{2} = 8 x .

Let

Δ_{1}

be the area of the triangle formed by the end points of its latus rectum and the point P(

\frac{1}{2}

,2) on the parabola and

Δ_{2}

be the area of the triangle formed by drawing tangents at P and at the end points of latus rectum.

\frac{Δ _{1}}{Δ _{2}}

is :

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Question Text	Consider the parabola $y^{2} = 8 x .$ Let $Δ_{1}$ be the area of the triangle formed by the end points of its latus rectum and the point P( $\frac{1}{2}$ ,2) on the parabola and $Δ_{2}$ be the area of the triangle formed by drawing tangents at P and at the end points of latus rectum. $\frac{Δ _{1}}{Δ _{2}}$ is :
Answer Type	Video solution: 1
Upvotes	150