A vertical line passing through the point (h, 0) intersects the ellipse x²/4 + y²/3 = 1 at the point

Q: Wait, why do we use the chord of contact formula? Can't we just find the equations of the tangents at P and Q and solve for their intersection?

You absolutely could, but that is the path of the 'brute force' student, not the 'JEE Advanced' student! Finding the tangent equations at $P$ and $Q$ involves slopes, point-slope forms, and tedious algebra. The chord of contact property $T=0$ is a shortcut that saves you time and reduces the probability of calculation errors. In competitive exams, elegance is efficiency.

Q: How do I know if the area function $\Delta(h)$ is increasing or decreasing without plotting it?

Calculus is your compass here! By finding the derivative $\Delta'(h)$, you can determine the sign of the slope. If $\Delta'(h) < 0$ for all $h$ in the interval $[1/2, 1]$, the function is strictly decreasing. This tells us immediately that the maximum occurs at the smallest $h$ and the minimum occurs at the largest $h$.

Q: Is the height of the triangle really just $\frac{4}{h} - h$? I'm struggling to visualize the geometry.

Let's re-orient. The vertical line is at $x=h$. The point $R$ lies on the x-axis at $x_R = 4/h$. Since the triangle $PQR$ is formed by the vertical line $x=h$ and the two tangents meeting at $R$, the height is simply the horizontal distance between the vertical line and the vertex $R$. That is $x_R - h$. You've got this!

Question

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h) = 	ext{area of the triangle } PQR$, $\Delta_1 = \max_{1/2 \le h \le 1} \Delta(h)$ and $\Delta_2 = \min_{1/2 \le h \le 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1 - 8\Delta_2 = $

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given an ellipse defined by the equation:

$$\frac{x^2}{4} + \frac{y^2}{3} = 1$$

We slice this ellipse with a vertical line $x=h$, where $h \in [\frac{1}{2}, 1]$. This slice creates two points, $P$ and $Q$. Tangents drawn at these points meet at a point $R$. Our objective is to analyze the area of the triangle $\Delta(h)$ formed by these three points.

### The Chord of Contact

To find the coordinates of $R(x_1, y_1)$, we utilize the **Chord of Contact** theorem. For any point $R(x_1, y_1)$ outside an ellipse, the line joining the points of tangency $P$ and $Q$ is given by the equation $T=0$:

$$\frac{xx_1}{4} + \frac{yy_1}{3} = 1$$

We are given that this line is $x=h$. By comparing the coefficients of the two equations, we deduce that $y_1 = 0$ and:

$$\frac{x_1}{4} = \frac{1}{h} \implies x_1 = \frac{4}{h}$$

Thus, the coordinates of the intersection point are $R(\frac{4}{h}, 0)$.

### Building the Area Function

The area of triangle $PQR$ is given by $\Delta(h) = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height}$.

First, we determine the base $PQ$. Substituting $x=h$ into the ellipse equation, we find $y = \pm \frac{\sqrt{3}}{2}\sqrt{4-h^2}$. The length of the base is:

$$	ext{Base} = 2 	imes \frac{\sqrt{3}}{2}\sqrt{4-h^2} = \sqrt{3}\sqrt{4-h^2}$$

Next, the height is the horizontal distance from $R(\frac{4}{h}, 0)$ to the line $x=h$:

$$	ext{Height} = \frac{4}{h} - h = \frac{4-h^2}{h}$$

Combining these, the area function is:

$$\Delta(h) = \frac{1}{2} \cdot \sqrt{3}\sqrt{4-h^2} \cdot \frac{4-h^2}{h} = \frac{\sqrt{3}(4-h^2)^{3/2}}{2h}$$

### The Calculus of Optimization

To find the extrema on the interval $[\frac{1}{2}, 1]$, we examine the derivative $\Delta'(h)$. Using the quotient rule, we obtain:

$$\Delta'(h) = -\sqrt{3} \left[ \frac{\sqrt{4-h^2}(h^2+2)}{h^2} ight]$$

Since $\Delta'(h) < 0$ for all $h$ in the given interval, the function is strictly decreasing. Consequently, the maximum value occurs at the lower bound $h = \frac{1}{2}$, and the minimum occurs at the upper bound $h = 1$.

Calculating these values:

$$\Delta_1 = \Delta\left(\frac{1}{2}ight) = \frac{\sqrt{3}(4 - 1/4)^{3/2}}{2(1/2)} = \frac{45\sqrt{5}}{8}$$

$$\Delta_2 = \Delta(1) = \frac{\sqrt{3}(4 - 1)^{3/2}}{2(1)} = \frac{9}{2}$$

### Final Calculation

We evaluate the final expression requested:

$$\frac{8}{\sqrt{5}} \Delta_1 - 8\Delta_2$$

Substituting our calculated values:

$$\frac{8}{\sqrt{5}} \left( \frac{45\sqrt{5}}{8} ight) - 8 \left( \frac{9}{2} ight) = 45 - 36 = 9$$

The final result is **9**.

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse and Line $x = h$

Intersection Points $P$ and $Q$

Finding Coordinates of $P$ and $Q$

Coordinates of $P$ and $Q$

Tangents and Point $R$

Chord of Contact

Coordinates of $R$

Triangle $P QR$

Area Formula for $Δ P QR$

Calculating Base and Height

Area Function $Δ (h)$

Maximizing and Minimizing Area

Derivative of Area Function

Calculating $Δ_{1}$ and $Δ_{2}$

Final Computation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Chord of Contact

Building the Area Function

The Calculus of Optimization

Final Calculation

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=h

Intersection Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Finding Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Tangents and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R