Let P(a sec θ, b tan θ) and Q(a sec φ, b tan φ), where θ + φ = π/2, be two points on the hyperbola x

Q: Wait, why do we reject the case where $\cos \theta = \sin \theta$?

That is a brilliant observation! If $\cos \theta = \sin \theta$, then $\theta = \frac{\pi}{4}$. Since $\phi = \frac{\pi}{2} - \theta$, this would mean $\phi = \frac{\pi}{4}$ as well. In this scenario, points $P$ and $Q$ are identical. If the points are the same, the normals are the same, and they don't intersect at a unique point $(h, k)$. We assume distinct points for the intersection to exist.

Q: Why can't we just use the slope-point form for the normal?

You absolutely could! However, in coordinate geometry, especially with conic sections, the parametric form is often the 'path of least resistance.' Using the slope-point form would force you to deal with derivatives and slopes, which introduces unnecessary complexity. The parametric form $ax \cos \theta + by \cot \theta = a^2 + b^2$ is specifically designed to handle the hyperbola's geometry elegantly.

Q: Is there a geometric interpretation of this result?

Yes! The result $k = -\frac{a^2 + b^2}{b}$ tells us that for any pair of points on the hyperbola whose eccentric angles sum to $\frac{\pi}{2}$, the y-coordinate of their intersection point is constant, regardless of the specific value of $\theta$. It is a beautiful property of the hyperbola's normal structure.

Question

Let $P(a \sec 	heta, b 	an 	heta)$ and $Q(a \sec \phi, b 	an \phi)$, where $	heta + \phi = \pi/2$, be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$, then $k$ is equal to

Options:

(A) $\frac{a^2+b^2}{a}$

(B) $-\left(\frac{a^2+b^2}{a}ight)$

(C) $\frac{a^2+b^2}{b}$

(D) $-\left(\frac{a^2+b^2}{b}ight)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the fascinating world of coordinate geometry. Today, we are uncovering a hidden symmetry within the hyperbola defined by the equation:

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

We consider two points, $P$ and $Q$, resting upon this curve. These points are defined by their eccentric angles, $	heta$ and $\phi$, respectively. We are given the critical constraint:

$$	heta + \phi = \frac{\pi}{2}$$

### The Normal Equation

To understand where these normals intersect, we must first define them. The normal to a hyperbola at a point $(a \sec 	heta, b 	an 	heta)$ is given by the equation:

$$ax \cos 	heta + by \cot 	heta = a^2 + b^2$$

This equation serves as the bedrock of our solution, dictating how the normal line behaves based on the angle $	heta$.

### The Symmetry of Complementary Angles

Now, let us examine point $Q$ with eccentric angle $\phi$. Using our standard formula, the normal at $Q$ is:

$$ax \cos \phi + by \cot \phi = a^2 + b^2$$

Since $\phi = \frac{\pi}{2} - 	heta$, we apply trigonometric identities: $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$. The normal at $Q$ transforms into:

$$ax \sin 	heta + by 	an 	heta = a^2 + b^2$$

Now, both normals are expressed in terms of the same variable, $	heta$. The complexity has vanished.

### The Algebraic Dance

We seek the intersection point $(h, k)$. Since this point lies on both lines, it must satisfy the following system:

1) $ah \cos 	heta + bk \cot 	heta = a^2 + b^2$

2) $ah \sin 	heta + bk 	an 	heta = a^2 + b^2$

To find $k$, we must eliminate $h$. We multiply the first equation by $\sin 	heta$ and the second by $\cos 	heta$:

$$ah \sin 	heta \cos 	heta + bk \cot 	heta \sin 	heta = (a^2 + b^2) \sin 	heta$$

$$ah \sin 	heta \cos 	heta + bk 	an 	heta \cos 	heta = (a^2 + b^2) \cos 	heta$$

Using the identities $\cot 	heta \sin 	heta = \cos 	heta$ and $	an 	heta \cos 	heta = \sin 	heta$, these simplify to:

$$ah \sin 	heta \cos 	heta + bk \cos 	heta = (a^2 + b^2) \sin 	heta$$

$$ah \sin 	heta \cos 	heta + bk \sin 	heta = (a^2 + b^2) \cos 	heta$$

### The Grand Cancellation

Subtracting the second equation from the first, the $ah \sin 	heta \cos 	heta$ terms cancel out perfectly:

$$bk(\cos 	heta - \sin 	heta) = (a^2 + b^2)(\sin 	heta - \cos 	heta)$$

We factor out the negative sign on the right side:

$$bk(\cos 	heta - \sin 	heta) = -(a^2 + b^2)(\cos 	heta - \sin 	heta)$$

Assuming $\cos 	heta 
eq \sin 	heta$, we divide both sides by $(\cos 	heta - \sin 	heta)$. The final result is:

**$$k = -\frac{a^2 + b^2}{b}$$**

Select Answer:

Visualized Solution

Hyperbola and Points $P, Q$

Equation of Normal

Normals at $P$ and $Q$

Transforming Normal at $Q$

Intersection Point $(h, k)$

Eliminating $h$ : Multiplication

Simplifying First Equation

Modifying Second Equation

Subtracting the Equations

Solving for $k$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Equation

The Symmetry of Complementary Angles

The Algebraic Dance

The Grand Cancellation

Similar Questions

Let $A (sec θ, 2 tan θ)$ and $B (sec ϕ, 2 tan ϕ)$ , where $θ + ϕ = π /2$ , be two points on the hyperbola $2 x^{2} - y^{2} = 2$ . If $(α, β)$ is the point of the intersection of the normals to the hyperbola at A and B, then $(2 β)^{2}$ is equal to .

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

Let $P (3, 3)$ be a point on the hyperbola, $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If the normal to it at $P$ intersects the x-axis at $(9, 0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to :

Let $P (6, 3)$ be a point on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If the normal at the point $P$ intersects the $x$ -axis at $(9, 0)$ , then the eccentricity of the hyperbola is

The point $P (- 26, 3)$ lies on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ having eccentricity $\frac{5}{2}$ . If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :

If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal at $P$ is:

Tangents are drawn to the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are

The normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{9} = 1$ at the point $(8, 33)$ on it passes through the point :

Select Answer:

Visualized Solution

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

Equation of Normal

Normals 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

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

Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,k)

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h: Multiplication