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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Parabola

JEE Advanced 2011 · Paper 1 · Mathematics · Parabola

Consider the parabola y^2=8 x . Let Δ_1 be the area of the triangle...

Consider the parabola y2=8xy^2=8 x. Let Δ1\Delta_1 be the area of the triangle formed by the end points of its latusrectum and the point $P(12,2)P\left(\frac{1}{2}, 2\right) on the parabola and Δ2\Delta_2 be the area of the triangle formed by drawing tangents at Pandattheendpointsofthelatusrectum.Then,P and at the end points of the latusrectum. Then,\frac{\Delta_1}{\Delta_2}$ isConsider the parabola y^2=8 x . Let Δ_1 be the area of the triangle formed by the end points of its latusrectum and the

Correct answer

2

Step-by-step solution

As, we know area of triangle formed by three points on parabola is twice the area of triangle formed by corresponding tangents, i.e. area of △PQR=2\triangle P Q R=2 area of ΔT1T2T3\Delta T_1 T_2 T_3. Consider the parabola y^2=8 x . Let Δ_1 be the area of the triangle formed by the end points of its latusrectum and the
∴Δ1=2Δ2\therefore \quad \Delta_1=2 \Delta_2 or Δ1Δ2=2\frac{\Delta_1}{\Delta_2}=2
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