If β is one of the angles between the normals to the ellipse, x²+3y² = 9 at the points (3 cosθ, √3 s

Q: Wait, why do we use the formula $m = \frac{a^2 y_1}{b^2 x_1}$ for the slope of the normal? Can't we just find the tangent slope and take the negative reciprocal?

You absolutely can! That is the fundamental definition of a normal. If you find the slope of the tangent $m_T$ by differentiating the ellipse equation, the slope of the normal is simply $-1/m_T$. The formula $m = \frac{a^2 y_1}{b^2 x_1}$ is just a shortcut derived from that exact process. Using the shortcut is faster, but understanding the derivation gives you the confidence to handle any curve!

Q: Why do we use the absolute value in the formula for $\tan \beta$?

Great question! When two lines intersect, they form two pairs of vertically opposite angles. One pair is acute, and the other is obtuse. The formula $\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ is designed to give you the tangent of the acute angle. Since the problem asks for 'the' angle $\beta$, we typically default to the acute one unless specified otherwise.

Q: Why does the final answer not depend on $\theta$?

It feels magical, doesn't it? In many JEE problems, especially those involving parametric coordinates on conics, the variable $\theta$ is designed to cancel out. It signifies that the geometric relationship—in this case, the angle between these specific normals—is a constant property of the ellipse, regardless of where you choose the points $P$ and $Q$!

Question

If $\beta$ is one of the angles between the normals to the ellipse, $x^2 + 3y^2 = 9$ at the points $(3 \cos 	heta, \sqrt{3} \sin 	heta)$ and $(-\sqrt{3} \sin 	heta, \sqrt{3} \cos 	heta)$; $	heta \in (0, \pi/2)$; then $\frac{2 \cot \beta}{\sin 2	heta}$ is equal to

Options:

(A) $1/\sqrt{3}$

(B) $2/\sqrt{3}$

(C) $\sqrt{2}$

(D) $\sqrt{3}/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a perfectly drawn ellipse, defined by the equation $x^2 + 3y^2 = 9$. It is not just a shape; it is a canvas of mathematical elegance.

To truly understand it, we must first bring it to its standard form. By dividing the entire equation by $9$, we get:

$$ \frac{x^2}{9} + \frac{y^2}{3} = 1 $$

Here, the semi-major axis $a$ is $3$, and the semi-minor axis $b$ is $\sqrt{3}$. This ellipse is stretched horizontally, a beautiful, flattened circle waiting for us to place our points.

### The Parametric Dance

We are given two points, $P(3 \cos 	heta, \sqrt{3} \sin 	heta)$ and $Q(-3 \sin 	heta, \sqrt{3} \cos 	heta)$. These are not random coordinates; they are parametric points that dance along the perimeter of our ellipse.

If you were to substitute them into our standard equation, you would see them fit perfectly. Our goal is to find the angle $\beta$ between the normals at these two points.

To do this, we need the slopes of these normals. The slope of a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $(x_1, y_1)$ is given by the powerful tool:

$$ m = \frac{a^2 y_1}{b^2 x_1} $$

### Calculating the Slopes

Let us apply this to point $P$. With $a^2 = 9$ and $b^2 = 3$, the slope $m_1$ becomes:

$$ m_1 = \frac{9(\sqrt{3} \sin 	heta)}{3(3 \cos 	heta)} = \sqrt{3} 	an 	heta $$

Now, for point $Q$, we repeat the process. The slope $m_2$ becomes:

$$ m_2 = \frac{9(\sqrt{3} \cos 	heta)}{3(-3 \sin 	heta)} = -\sqrt{3} \cot 	heta $$

We have our two slopes, and the stage is set.

### The Angle Between Normals

To find the angle $\beta$ between these two lines, we use the classic formula:

$$ 	an \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| $$

Substituting our slopes, we get:

$$ 	an \beta = \left| \frac{\sqrt{3} 	an 	heta - (-\sqrt{3} \cot 	heta)}{1 + (\sqrt{3} 	an 	heta)(-\sqrt{3} \cot 	heta)} ight| $$

Simplifying the numerator, we factor out $\sqrt{3}$ to get $\sqrt{3}(	an 	heta + \cot 	heta)$. In the denominator, the product $	an 	heta \cot 	heta$ is simply $1$, so we have $1 - 3$, which is $-2$. Because of the absolute value, the negative sign vanishes, leaving us with:

$$ 	an \beta = \frac{\sqrt{3}(	an 	heta + \cot 	heta)}{2} $$

### The Final Calculation

We know that $	an 	heta + \cot 	heta = \frac{1}{\sin 	heta \cos 	heta}$. Using the double angle identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$, we can write this as $\frac{2}{\sin 2	heta}$.

Substituting this back into our expression for $	an \beta$, we get:

$$ 	an \beta = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sin 2	heta} = \frac{\sqrt{3}}{\sin 2	heta} $$

This implies that $\cot \beta = \frac{\sin 2	heta}{\sqrt{3}}$. Finally, we calculate the requested value:

$$ \frac{2 \cot \beta}{\sin 2	heta} = \frac{2 (\sin 2	heta / \sqrt{3})}{\sin 2	heta} $$

The $\sin 2	heta$ terms cancel out, leaving us with the elegant result of **$\frac{2}{\sqrt{3}}$**.

Animated Solution for Mathematics - Conic Sections: If $β$ is one of the angles between the normals to the ellipse, $x^{2} + 3 y^{2} = 9$ at the points $(3 cos θ, 3 sin θ)$ and $(- 3 sin θ, 3 cos θ)$ ; $θ \in (0, π /2)$ ; then $\frac{2 c o t β}{s i n 2 θ}$ is equal to

Select Answer:

Visualized Solution

Visualize the Ellipse

Identify Points $P$ and $Q$

Slope of Normal Formula

Slope of Normal at $P$

Slope of Normal at $Q$

Angle Between Normals

Substitute Slopes into Angle Formula

Simplify the Denominator

Simplify the Numerator

Apply Double Angle Formula

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

Calculating the Slopes

The Angle Between Normals

The Final Calculation

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