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MathHardAdv 2013 • Paper 1

Area of Triangle Formed by Ellipse Tangents and Chord

Suppose a vertical line through the point (h,0)(h, 0) meets the ellipse x24+y23=1\frac{x^2}{4} + \frac{y^2}{3} = 1 at points PP and QQ. Let the tangents drawn to the ellipse at PP and QQ meet at the point RR. If Δ(h)\Delta(h) represents the area of the triangle PQRPQR, with Δ1=max⁡1/2≤h≤1Δ(h)\Delta_1 = \max_{1/2 \le h \le 1} \Delta(h) and Δ2=min⁡1/2≤h≤1Δ(h)\Delta_2 = \min_{1/2 \le h \le 1} \Delta(h), determine the value of 85Δ1−8Δ2\frac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2

Official Numerical Answer9

Topics & Concepts

Conic SectionsEllipseTangents and NormalsMaxima and Minima

Step-by-Step Solution

To find the value of 85Δ1−8Δ2\frac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2, we follow these steps:

Step 1: Find the coordinates of PP, QQ, and RR

The equation of the ellipse is given by: x24+y23=1\frac{x^2}{4} + \frac{y^2}{3} = 1

A vertical line through (h,0)(h, 0) is represented by x=hx = h. Substituting x=hx = h into the equation of the ellipse: h24+y23=1  ⟹  y2=3(1−h24)=34(4−h2)\frac{h^2}{4} + \frac{y^2}{3} = 1 \implies y^2 = 3\left(1 - \frac{h^2}{4}\right) = \frac{3}{4}(4 - h^2)

Thus, the coordinates of the intersection points PP and QQ are: P=(h,324−h2)andQ=(h,−324−h2)P = \left(h, \frac{\sqrt{3}}{2}\sqrt{4 - h^2}\right) \quad \text{and} \quad Q = \left(h, -\frac{\sqrt{3}}{2}\sqrt{4 - h^2}\right)

The line PQPQ serves as the chord of contact of the tangents drawn from the point of intersection R(x1,y1)R(x_1, y_1) to the ellipse. The equation of the chord of contact from R(x1,y1)R(x_1, y_1) is: xx14+yy13=1\frac{x x_1}{4} + \frac{y y_1}{3} = 1

Comparing this with the equation of the line PQPQ, which is x=hx = h (or xh=1\frac{x}{h} = 1): x14=1h  ⟹  x1=4handy1=0\frac{x_1}{4} = \frac{1}{h} \implies x_1 = \frac{4}{h} \quad \text{and} \quad y_1 = 0

Hence, the point of intersection is: R=(4h,0)R = \left(\frac{4}{h}, 0\right)

Step 2: Determine the area of triangle PQRPQR

The chord PQPQ lies on the vertical line x=hx = h. Therefore, the base of △PQR\triangle PQR is the segment PQPQ: Base=∣yP−yQ∣=2⋅324−h2=34−h2\text{Base} = |y_P - y_Q| = 2 \cdot \frac{\sqrt{3}}{2}\sqrt{4 - h^2} = \sqrt{3}\sqrt{4 - h^2}

The altitude from the vertex R(4h,0)R\left(\frac{4}{h}, 0\right) to the line x=hx = h is: Height=∣4h−h∣=4−h2h(since h∈[1/2,1]  ⟹  4h>h)\text{Height} = \left|\frac{4}{h} - h\right| = \frac{4 - h^2}{h} \quad (\text{since } h \in [1/2, 1] \implies \frac{4}{h} > h)

Thus, the area of △PQR\triangle PQR, denoted by Δ(h)\Delta(h), is: Δ(h)=12×Base×Height=12⋅34−h2⋅4−h2h=32(4−h2)3/2h\Delta(h) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \cdot \sqrt{3}\sqrt{4 - h^2} \cdot \frac{4 - h^2}{h} = \frac{\sqrt{3}}{2} \frac{(4 - h^2)^{3/2}}{h}

Step 3: Find the maximum and minimum of Δ(h)\Delta(h) on [1/2,1][1/2, 1]

Let f(h)=(4−h2)3/2hf(h) = \frac{(4 - h^2)^{3/2}}{h}. Differentiating with respect to hh: f′(h)=h⋅32(4−h2)1/2(−2h)−(4−h2)3/2⋅1h2f'(h) = \frac{h \cdot \frac{3}{2}(4 - h^2)^{1/2}(-2h) - (4 - h^2)^{3/2} \cdot 1}{h^2} f′(h)=4−h2[−3h2−(4−h2)]h2=−4−h2(2h2+4)h2f'(h) = \frac{\sqrt{4 - h^2}\left[-3h^2 - (4 - h^2)\right]}{h^2} = -\frac{\sqrt{4 - h^2}(2h^2 + 4)}{h^2}

Since 2h2+4>02h^2 + 4 > 0 and 4−h2>0\sqrt{4 - h^2} > 0 for all h∈[1/2,1]h \in [1/2, 1], we have: f′(h)<0for all h∈[1/2,1]f'(h) < 0 \quad \text{for all } h \in [1/2, 1]

This implies that Δ(h)\Delta(h) is a strictly decreasing function on the interval [12,1]\left[\frac{1}{2}, 1\right]. Therefore:

  • The maximum value occurs at h=12h = \frac{1}{2}: Δ1=Δ(12)=32(4−14)3/212=3(154)3/2=3⋅15158=4558\Delta_1 = \Delta\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2} \frac{\left(4 - \frac{1}{4}\right)^{3/2}}{\frac{1}{2}} = \sqrt{3} \left(\frac{15}{4}\right)^{3/2} = \sqrt{3} \cdot \frac{15\sqrt{15}}{8} = \frac{45\sqrt{5}}{8}

  • The minimum value occurs at h=1h = 1: Δ2=Δ(1)=32(4−1)3/21=32⋅33=92\Delta_2 = \Delta(1) = \frac{\sqrt{3}}{2} \frac{(4 - 1)^{3/2}}{1} = \frac{\sqrt{3}}{2} \cdot 3\sqrt{3} = \frac{9}{2}

Step 4: Calculate the required expression

Substituting the values of Δ1\Delta_1 and Δ2\Delta_2: 85Δ1−8Δ2=85(4558)−8(92)=45−36=9\frac{8}{\sqrt{5}}\Delta_1 - 8\Delta_2 = \frac{8}{\sqrt{5}}\left(\frac{45\sqrt{5}}{8}\right) - 8\left(\frac{9}{2}\right) = 45 - 36 = 9

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