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

Q: Wait, why do the tangents meet on the x-axis?

Because the ellipse is symmetric about the x-axis, the tangent at $(x, y)$ and the tangent at $(x, -y)$ must be reflections of each other across the x-axis. Any two non-parallel lines that are reflections of each other across an axis must intersect on that axis. It is a beautiful geometric consequence of symmetry!

Q: Can I use parametric coordinates instead of $(h, y)$?

Absolutely! You could use $x = 2\cos\theta$ and $y = \sqrt{3}\sin\theta$. However, since the problem gives us the vertical line $x=h$, working directly with $h$ is often faster and less prone to trigonometric identity errors. Both methods will lead you to the same elegant result.

Q: How do I know the area is decreasing without finding the derivative?

You can test the endpoints. At $h=1/2$, the chord $PQ$ is longer and the intersection $R$ is further away, creating a larger triangle. As $h$ approaches 1, the chord $PQ$ shrinks to a point, and the triangle collapses, so the area must decrease. Calculus just confirms this intuition rigorously!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The ellipse is defined by the equation:

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

A vertical line $x=h$ intersects the ellipse at points $P$ and $Q$. We are interested in the area of the triangle $PQR$, where $R$ is the intersection point of the tangents to the ellipse at $P$ and $Q$.

### Finding the Intersection

Substituting $x=h$ into the ellipse equation yields $y^2 = \frac{3}{4}(4-h^2)$. Thus, the coordinates of the points are:

$$P = \left(h, \frac{\sqrt{3}}{2}\sqrt{4-h^2}ight), \quad Q = \left(h, -\frac{\sqrt{3}}{2}\sqrt{4-h^2}ight)$$

The equation of the tangent at a point $(x_1, y_1)$ on the ellipse is given by $\frac{xx_1}{4} + \frac{yy_1}{3} = 1$. For point $P$, this becomes:

$$\frac{xh}{4} + \frac{y\left(\frac{\sqrt{3}}{2}\sqrt{4-h^2}ight)}{3} = 1$$

Due to the symmetry of the ellipse, the tangents at $P$ and $Q$ intersect on the $x$-axis. Setting $y=0$ in the tangent equation, we find $\frac{xh}{4} = 1$, which identifies the intersection point $R$ as:

$$R = \left(\frac{4}{h}, 0ight)$$

### Constructing the Area Function

The area of triangle $PQR$ is given by $\Delta(h) = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. The base $PQ$ is the vertical distance between $P$ and $Q$:

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

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

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

Combining these, the area function $\Delta(h)$ is:

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

### The Calculus of Change

To analyze the behavior of the area, we examine $f(h) = \frac{(4-h^2)^{3/2}}{h}$. Applying the quotient rule, the derivative is:

$$f'(h) = \frac{-(4-h^2)^{1/2}(2h^2+4)}{h^2}$$

Since $h \in [1/2, 1]$, the derivative $f'(h)$ is strictly negative. This indicates that the area function is strictly decreasing as $h$ increases.

### Final Calculation

The maximum area $\Delta_1$ occurs at the lower bound $h=1/2$:

$$\Delta_1 = \frac{\sqrt{3}}{2} \frac{(4 - 1/4)^{3/2}}{1/2} = \sqrt{3} \left(\frac{15}{4}ight)^{3/2} = \frac{45\sqrt{5}}{8}$$

The minimum area $\Delta_2$ occurs at the upper bound $h=1$:

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

Substituting these values into the target expression $\frac{8}{\sqrt{5}} \Delta_1 - 8\Delta_2$:

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

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse and Vertical Line $x = h$

Finding Coordinates of $P$ and $Q$

Tangents at $P$ and $Q$

Finding Intersection Point $R$

Setting up the Area of Triangle $P QR$

Expressing Area $Δ (h)$

Analyzing Monotonicity of $Δ (h)$

Calculating $Δ_{1}$ and $Δ_{2}$

Final Computation of $\frac{8}{5} Δ_{1} - 8 Δ_{2}$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Finding the Intersection

Constructing the Area Function

The Calculus of Change

Final Calculation

Similar Questions

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Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

Enter Numerical Value:

Visualized Solution

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

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

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

Setting up the Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQR

Expressing Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ(h)

Analyzing Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ(h)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ2​

Final Computation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5​8​Δ1​−8Δ2​