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 \left(\right. \frac{1}{2} , \textrm{ }\textrm{ } 2 \left.\right) on the parabola, and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then \frac{\left(\Delta\right)_{1}}{\left(\Delta\right)

Engineering

Mathematics

Highlights on Parabola

Question

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 $P (\frac{1}{2}, 2)$ on the parabola, and Δ₂ be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then $\frac{Δ_{1}}{Δ_{2}}$ is

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It is a property that area of triangle formed by joining three points lying on parabola is twice the area of triangle formed by tangents at these points

Alternate : y² = 8x

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

$Δ_{1} = \frac{1}{2} | Base \times Height |$

$= \frac{1}{2} \times \frac{3}{2} \times 8 = 6$

Also

Equation of tangent at $P (\frac{1}{2}, 2)$

$y (2) = 4 . (x + \frac{1}{2})$

y = 2x + 1 ….(1)

Tangent at A : y = x + 2

Tangent at B : – y = + x + 2 ⇒ y = – x – 2

Point of intersection

L(–2, 0), M (1, 3), N (–1, –1)

$Δ_{2} = | \frac{1}{2} | \begin{matrix} - 2 & 0 & 1 \\ 1 & 3 & 1 \\ - 1 & - 1 & 1 \end{matrix} | |$

$= | \frac{1}{2} [- 2 (4) + (- 1 + 3)] |$

$= | \frac{1}{2} [- 8 + 3 - 1] | = 3$

$So, \frac{Δ_{1}}{Δ_{2}} = \frac{6}{3} = 2$