IIT-JEE 2011 · Paper 1 · Mathematics Q63

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MathematicsInteger+4Question 63 of 69

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 $P\left(\frac12,2\right)$ 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 $\dfrac{\Delta_1}{\Delta_2}$ is

Your answer

Answer and worked solution

Official answer: 2

Δ₁. Latus rectum ends $(2,\pm4)$ and $P\left(\frac12,2\right)$ : base 8 on $x=2$ , height $\frac32$ , area 6.

Δ₂. For a parabola, the triangle formed by the tangents at three points has half the area of the triangle formed by the points themselves: $\Delta_2=3$ . (Directly: the tangents meet at $(-2,0)$ , $(-1,1)$ and $(-1,-3)$ , area 3.)

Ratio 2.

Answer: 2

Topics:Parabola Tangents and normals to conics

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IIT-JEE 2011 · Paper 1, Mathematics question 63