Consider the parabola $y^{2} = 8x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P\left(\frac{1}{2}, 2\right)$ on the parabola and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{\Delta_1}{\Delta_2}$ is

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(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 Δ₁/Δ₂ is $\text{Given: } y^2 = 8x$ $P\left(\dfrac{1}{2},\, 2\right) \text{ lies on the parabola}$ $\Delta_1:\ \triangle \text{ with vertices = latus-rectum ends and } P$ $\Delta_2:\ \triangle \text{ formed by tangents at } P \text{ and the latus-rectum ends}$ $\text{To find: } \dfrac{\Delta_1}{\Delta_2}$ $\text{Plan:}\quad 1.\ \text{find } a \text{ and the latus rectum}$ $2.\ \Delta_1 = \text{area of } \triangle ABP$ $3.\ \text{tangents at } P, A, B \Rightarrow \Delta_2 \Rightarrow \dfrac{\Delta_1}{\Delta_2}$ $y^2 = 4ax \Rightarrow 4a = 8 \Rightarrow a = 2$ $\text{Focus} = (a,0) = (2,0)$ $\text{Parabola } y^2 = 8x \text{ opens right}$ $\text{Latus rectum: } x = a = 2,\quad \text{length } 4a = 8$ $A(2,4),\qquad B(2,-4)$ $P\left(\dfrac{1}{2},\, 2\right) \text{ is given on the parabola}$ $\Delta_1 = [\triangle ABP]$ $AB = 8,\qquad h = 2 - \dfrac{1}{2} = \dfrac{3}{2}$ $\Delta_1 = \dfrac{1}{2}\cdot 8 \cdot \dfrac{3}{2} = 6$ $\text{Tangent to } y^2 = 4ax \text{ at } (x_1,y_1):\ yy_1 = 2a(x+x_1)$ $\text{At } P:\ 2y = 4\left(x+\dfrac{1}{2}\right) \Rightarrow y = 2x+1$ $\text{At } A(2,4):\ y = x+2$ $\text{At } B(2,-4):\ y = -x-2$ $U = (-2,0) \text{ from } y=x+2,\ y=-x-2$ $V = (1,3) \text{ from } y=2x+1,\ y=x+2$ $W = (-1,-1) \text{ from } y=2x+1,\ y=-x-2$ $\Delta_2 = [\triangle UVW]$ $\Delta_2 = \dfrac{1}{2}\left|(-2)(3+1) + 1(-1-0) + (-1)(0-3)\right|$ $= \dfrac{1}{2}\,|-8-1+3| = 3$ $\dfrac{\Delta_1}{\Delta_2} = \dfrac{6}{3} = 2$ $\boxed{2}$