Consider the parabola . Let be the area of the triangle formed by the end points of its latus rectum and the point on the parabola and be the area of the triangle formed by drawing tangents at and at the end points of the latus rectum. Then is

Question

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 the latus rectum. Then $\frac{Δ_{1}}{Δ_{2}}$ is [2011]

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(2)

$Δ_{1} = area (∆ P L L') = \frac{1}{2} \times 8 \times \frac{3}{2} = 6$

$Equation of AB is y = 2 x + 1; equation of AC is y = x + 2 and equation of BC is - y = x + 2 .$

On solving the above equations pairwise, we get $A (1, 3), B (- 1, - 1) and C (- 2, 0)$

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

$∴$ $\frac{Δ_{1}}{Δ_{2}} = 2$

Solution

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Solution

Q. 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(12,2) 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 Δ1Δ2 is [2011]

Ans. (2) Δ1=area(∆PLL')=12×8×32=6 Equation of AB is y=2x+1; equation of AC is y=x+2 and equation of BC is -y=x+2. On solving the above equations pairwise, we get A(1,3), B(-1,-1) and C(-2,0) ∴ Δ2=12|131-1-11-201|=3 ∴ Δ1Δ2=2

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