Consider a hyperbola H : x² - 2y² = 4. Let the tangent at a point P(4, √6) meet the x-axis at Q and

Q: Wait, why do we choose the focus $F(\sqrt{6}, 0)$ and not $(-\sqrt{6}, 0)$?

That is a fantastic observation! The hyperbola is symmetric, so it has two foci, one on each branch. The problem explicitly states that $F$ is the focus 'nearer to the point $P$'. Since $P(4, \sqrt{6})$ has a positive x-coordinate, it sits on the right branch of the hyperbola. Therefore, the focus on the positive x-axis is the one that is physically closer to $P$. Always let the geometry guide your choice of coordinates!

Q: How can we be so sure that $\Delta QFR$ is a right-angled triangle?

Great question! In coordinate geometry, we don't guess; we prove. We found that $Q$ lies on the x-axis, and $F$ also lies on the x-axis, so the segment $QF$ is horizontal. The latus rectum, by definition, is a line perpendicular to the transverse axis (the x-axis) passing through the focus. Since $R$ lies on the latus rectum, the segment $FR$ is vertical. A horizontal line and a vertical line are perpendicular by definition. Thus, the angle at $F$ is $90^\circ$.

Q: Is there a faster way to find the area without calculating the coordinates of $R$?

While there are properties involving the tangent and the latus rectum, calculating the coordinates is the most robust and error-proof method for JEE Advanced. It builds your confidence and ensures you don't miss a sign or a factor. The algebra here is quite clean, so sticking to the coordinate geometry approach is actually the most efficient path to the correct answer.

Question

Consider a hyperbola $H : x^2 - 2y^2 = 4$. Let the tangent at a point $P(4, \sqrt{6})$ meet the $x$-axis at $Q$ and latus rectum at $R(x_1, y_1), x_1 > 0$. If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to

Options:

(A) $4\sqrt{6}$

(B) $\sqrt{6} - 1$

(C) $\frac{7}{\sqrt{6}} - 2$

(D) $4\sqrt{6} - 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on this mathematical journey. Today, we explore the elegant architecture of a hyperbola, which acts as a gateway defining the limits of space in our coordinate system.

Our given equation is $x^2 - 2y^2 = 4$. To navigate this map, we must first bring it into its standard form by dividing the entire equation by $4$:

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

Here, we identify the parameters of our hyperbola: $a^2 = 4$ and $b^2 = 2$. This is the foundation upon which we will build our entire solution.

### Finding the Focus: The Anchor Point

Every hyperbola is defined by its foci—the points that act as the gravitational centers of the curve. To find them, we calculate the eccentricity $e$ using the relationship:

$$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$$

The distance of the focus from the center is given by $ae$. With $a = 2$, we find:

$$ae = 2 	imes \sqrt{\frac{3}{2}} = \sqrt{6}$$

Thus, our foci are located at $(\pm \sqrt{6}, 0)$. Given point $P(4, \sqrt{6})$ sits on the right branch of the hyperbola, the focus 'nearer' to $P$ is the one on the positive x-axis: $F(\sqrt{6}, 0)$.

### The Tangent's Path

Now, we determine the tangent line at point $P(x_1, y_1)$ using the formula:

$$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$

Substituting our values $a^2=4$, $b^2=2$, $x_1=4$, and $y_1=\sqrt{6}$, we obtain:

$$\frac{4x}{4} - \frac{\sqrt{6}y}{2} = 1 \Rightarrow 2x - \sqrt{6}y = 2$$

This line intersects the x-axis at point $Q$. Setting $y=0$ in the tangent equation, we find $2x = 2$, which gives $x=1$. Thus, $Q$ is $(1, 0)$.

### The Latus Rectum and the Triangle

The latus rectum is a vertical line passing through the focus $F(\sqrt{6}, 0)$, defined by the equation $x = \sqrt{6}$. To find point $R$, where our tangent meets this vertical line, we substitute $x = \sqrt{6}$ into the tangent equation:

$$2(\sqrt{6}) - \sqrt{6}y = 2 \Rightarrow \sqrt{6}y = 2\sqrt{6} - 2$$

Solving for $y$, we get $y = 2 - \frac{2}{\sqrt{6}}$. Therefore, our point $R$ is $(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$.

### The Final Triumph: Calculating the Area

We now consider the triangle $\Delta QFR$ with vertices $Q(1, 0)$, $F(\sqrt{6}, 0)$, and $R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$. Since $QF$ lies on the x-axis and $FR$ is a vertical segment, this is a right-angled triangle at $F$.

The area is calculated as $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. The base $QF$ is $\sqrt{6} - 1$, and the height $FR$ is $2 - \frac{2}{\sqrt{6}}$.

$$	ext{Area} = \frac{1}{2} 	imes (\sqrt{6} - 1) 	imes 2\left(1 - \frac{1}{\sqrt{6}}ight)$$

The $2$ and the $\frac{1}{2}$ cancel out, leaving:

$$	ext{Area} = (\sqrt{6} - 1)\left(1 - \frac{1}{\sqrt{6}}ight) = \sqrt{6} - 1 - 1 + \frac{1}{\sqrt{6}} = \sqrt{6} - 2 + \frac{1}{\sqrt{6}}$$

Combining these terms, we arrive at the final result:

$$\mathbf{	ext{Area} = \frac{7}{\sqrt{6}} - 2}$$

Select Answer:

Visualized Solution

Standard Form of Hyperbola

Eccentricity and Foci

Focus Nearer to $P$

Equation of Tangent at $P$

Finding Point $Q$

Equation of Latus Rectum at $F$

Finding Point $R$

Geometry of $Δ QF R$

Calculating Base $QF$

Calculating Height $F R$

Area of Triangle $QF R$

Final Simplification

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Finding the Focus

The Tangent's Path

The Latus Rectum and the Triangle

The Final Triumph

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Select Answer:

Visualized Solution

Standard Form of Hyperbola

Eccentricity and Foci

Focus Nearer to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Equation of Latus Rectum at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F

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

Geometry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔQFR

Calculating Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QF

Calculating Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FR

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