Consider the hyperbola x²/a² - y²/b² = 1 having one of its focus at P(-3, 0). If the latus ractum th

Q: Wait, why is the other focus at $(3, 0)$? Is it always symmetric?

Yes, absolutely! The standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is centered at the origin $(0, 0)$. Since the foci are at $(\pm ae, 0)$, if one focus is at $(-3, 0)$, the other must be at $(3, 0)$ by the inherent symmetry of the coordinate axes. This is a fundamental property you should always look for!

Q: Why can't we just use the distance formula instead of slopes for the right angle?

You certainly can! If $\angle LPL' = 90^\circ$, then by the Pythagorean theorem, $PL^2 + PL'^2 = LL'^2$. However, using slopes is often much faster in coordinate geometry because the condition $m_1 \cdot m_2 = -1$ is algebraically simpler than dealing with square roots in the distance formula. Both methods will lead you to the exact same result.

Q: Why do we reject the negative root for $a$?

Great catch! In the context of the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $a$ represents the semi-major axis, which is a physical length. Lengths must always be positive. When we solve the quadratic equation $a^2 + 6a - 9 = 0$, we get two roots, but only the positive one makes physical sense for the geometry of the hyperbola.

Q: How do we know the latus rectum is a vertical line?

By definition, the latus rectum of a hyperbola is a chord passing through a focus and perpendicular to the transverse axis. Since our transverse axis is the x-axis, any line perpendicular to it must be a vertical line of the form $x = k$. Since it passes through the focus $S(3, 0)$, the equation of the latus rectum is simply $x = 3$.

Question

Consider the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having one of its focus at $P(-3, 0)$. If the latus ractum through its other focus subtends a right angle at $P$ and $a^{2}b^{2} = \alpha\sqrt{2} - \beta, \alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a beautiful hyperbola problem.

Imagine you are standing on the Cartesian plane, looking at the hyperbola defined by:

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

We are given one focus at $P(-3, 0)$. Because the hyperbola is perfectly symmetric about the y-axis, we immediately know the other focus, $S$, must be at $(3, 0)$.

This gives us our first vital clue: the distance from the center to the focus is $ae = 3$.

### The Latus Rectum's Secret

Now, let us focus on the latus rectum. It is a vertical line passing through the focus $S(3, 0)$.

The endpoints of this latus rectum, $L$ and $L'$, are the points where this vertical line intersects the hyperbola. Using the standard formula for the latus rectum of a hyperbola, the y-coordinates are $\frac{b^2}{a}$ and $-\frac{b^2}{a}$.

So, our points are $L(3, \frac{b^2}{a})$ and $L'(3, -\frac{b^2}{a})$.

### The Right Angle Condition

The problem drops a fascinating hint: the latus rectum subtends a right angle at the focus $P(-3, 0)$. In the language of coordinate geometry, this means the product of the slopes of lines $PL$ and $PL'$ must be $-1$.

Let us calculate these slopes:

$$m_{PL} = \frac{\frac{b^2}{a} - 0}{3 - (-3)} = \frac{b^2}{6a}$$

By symmetry, the slope of $PL'$ is simply the negative of this:

$$m_{PL'} = -\frac{b^2}{6a}$$

### The Algebraic Dance

Now, we set the product of these slopes to $-1$:

$$\left( \frac{b^2}{6a} \right) \cdot \left( -\frac{b^2}{6a} \right) = -1$$

This simplifies to $-\frac{b^4}{36a^2} = -1$, which gives us the beautiful relation $b^4 = 36a^2$, or simply $b^2 = 6a$.

We recall the fundamental identity for a hyperbola: $b^2 = a^2(e^2 - 1)$. Expanding this, we get $b^2 = a^2e^2 - a^2$. Since $ae = 3$, then $a^2e^2 = 9$, leading to $b^2 = 9 - a^2$.

### Solving for the Unknowns

We now have a system of two equations: $b^2 = 6a$ and $b^2 = 9 - a^2$. Equating them, we get:

$$6a = 9 - a^2 \Rightarrow a^2 + 6a - 9 = 0$$

Using the quadratic formula, we find $a = -3 \pm 3\sqrt{2}$. Since $a$ must be positive, we choose:

$$a = 3\sqrt{2} - 3 = 3(\sqrt{2} - 1)$$

### The Final Stretch

The question asks for $a^2b^2$. Since $b^2 = 6a$, then $a^2b^2 = a^2(6a) = 6a^3$.

Substituting our value for $a$:

$$6[3(\sqrt{2} - 1)]^3 = 162(\sqrt{2} - 1)^3$$

Expanding $(\sqrt{2} - 1)^3$ using the binomial expansion, we get $2\sqrt{2} - 6 + 3\sqrt{2} - 1 = 5\sqrt{2} - 7$. Multiplying by $162$:

$$162(5\sqrt{2} - 7) = 810\sqrt{2} - 1134$$

Comparing this to $\alpha\sqrt{2} - \beta$, we find $\alpha = 810$ and $\beta = 1134$. The sum **$\alpha + \beta = 1944$**. A truly satisfying conclusion to a wonderful journey!

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola

The Latus Rectum

The Right Angle Condition

Calculating the Slopes

Relating $a$ and $b$

Using the Eccentricity Relation

Solving for $a$

Selecting the Valid Root

Setting up $a^{2} b^{2}$

Expanding the Cubic Expression

Final Values of $α$ and $β$

The Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Latus Rectum's Secret

The Right Angle Condition

The Algebraic Dance

Solving for the Unknowns

The Final Stretch

Similar Questions

Let the latus rectum of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + n)$ , where $l, m$ and $n$ are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to ______

Let the ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$ and the hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$ , then the value of $24 (e + L)$ is :

Consider a branch of the hyperbola $x^{2} - 2 y^{2} - 22 x - 42 y - 6 = 0$ with vertex at the point $A$ . Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$ , then the area of the triangle $A B C$ is

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

If the equation of the hyperbola with foci $(4, 2)$ and $(8, 2)$ is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to _____.

Let the foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola

The Latus Rectum

The Right Angle Condition

Calculating the Slopes

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Using the Eccentricity Relation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Selecting the Valid Root

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2b2

Expanding the Cubic Expression

Final Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

The Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Latus Rectum's Secret

The Right Angle Condition

The Algebraic Dance

Solving for the Unknowns

The Final Stretch

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of the ellipse 16x2​+7y2​=1 and the hyperbola 144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is:-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the equation of the hyperbola with foci (4,2) and (8,2) is 3x2−y2−αx+βy+γ=0, then α+β+γ is equal to _____.

Relating $a$ and $b$

Solving for $a$

Setting up $a^{2} b^{2}$

Final Values of $α$ and $β$

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

If the equation of the hyperbola with foci $(4, 2)$ and $(8, 2)$ is $3 x^{2} - y^{2} - α x + β y + γ = 0$ , then $α + β + γ$ is equal to _____.