An angle of intersection of the curves, x²/a² + y²/b² = 1 and x² + y² = ab, a b, is : (1) tan⁻¹((a

Q: Why do we bother finding the ratio $\frac{x_1}{y_1}$ instead of solving for $x_1$ and $y_1$ individually? It feels like we are skipping steps.

That is a brilliant observation! In JEE problems, solving for individual coordinates often leads to terrifying radicals and complex expressions that are prone to calculation errors. By focusing on the ratio $\frac{x_1}{y_1}$, we exploit the symmetry of the equations. The variables $x_1$ and $y_1$ are just placeholders for the slope; the ratio is the true geometric essence. Always look for the 'invariant' or the ratio that simplifies the expression!

Q: Wait, why do we use the absolute value in the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$?

Great question! When two curves intersect, they actually create two angles: an acute one and an obtuse one (which sum to $180^\circ$). By convention, the 'angle of intersection' refers to the acute angle. The absolute value ensures that our result for $\tan \theta$ is positive, giving us the acute angle directly. If we didn't use it, we might get a negative value, which would just indicate the obtuse angle.

Q: Is there a way to solve this without calculus? Maybe using coordinate geometry properties?

While you could potentially use properties of ellipses and circles, calculus is the 'universal key.' It works regardless of the curve's complexity. In the JEE Advanced exam, you want a method that is robust and reliable. Implicit differentiation is that method. It turns a geometric problem into a systematic algebraic one, which is exactly what you want under pressure.

Question

An angle of intersection of the curves, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $x^2 + y^2 = ab, a > b$, is :

Options:

(A) $	an^{-1}\left(\frac{a+b}{\sqrt{ab}}ight)$

(B) $	an^{-1}\left(\frac{a-b}{2\sqrt{ab}}ight)$

(C) $	an^{-1}\left(\frac{a-b}{\sqrt{ab}}ight)$

(D) $	an^{-1}(2\sqrt{ab})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a vast coordinate plane. Before you lie two distinct paths: an ellipse, defined by

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

and a circle, defined by

$$ x^2 + y^2 = ab $$

They are not parallel; they cross. Our mission is to find the angle at which they meet. This is an exploration of how two different geometric entities share a single point in space.

### The Calculus of Slopes

To find the angle of intersection, we need to know the direction of each curve at the point of intersection, $(x_1, y_1)$. We utilize implicit differentiation to find the slopes.

For the ellipse, we differentiate

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

with respect to $x$, yielding

$$ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 $$

Solving for the slope $m_1$, we get

$$ m_1 = -\frac{b^2 x_1}{a^2 y_1} $$

Similarly, for the circle $x^2 + y^2 = ab$, the derivative is $2x + 2y \frac{dy}{dx} = 0$, giving us

$$ m_2 = -\frac{x_1}{y_1} $$

These slopes represent the instantaneous direction of the curves at the point of intersection.

### The Algebraic Dance

We now face the challenge of the unknown coordinates $(x_1, y_1)$. By substituting $y_1^2 = ab - x_1^2$ into the ellipse equation, we can isolate $x_1^2$.

After algebraic maneuvering, we find

$$ x_1^2 = \frac{a^2 b}{a + b} \quad 	ext{and} \quad y_1^2 = \frac{ab^2}{a + b} $$

The magic happens when we look at the ratio:

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

This ratio is the key that unlocks the entire problem. We do not need the individual coordinates; we only need their relationship.

### The Final Convergence

With the ratio

$$ \frac{x_1}{y_1} = \sqrt{\frac{a}{b}} $$

in our pocket, we turn to the angle formula:

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

Substituting our slopes $m_1 = -\frac{b^2 x_1}{a^2 y_1}$ and $m_2 = -\frac{x_1}{y_1}$, the expression simplifies significantly.

The numerator becomes

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

and the denominator simplifies to

$$ \frac{a + b}{a} $$

When we combine them, the $(a+b)$ terms vanish, leaving us with the elegant result:

$$ 	an 	heta = \frac{a - b}{\sqrt{ab}} $$

This is the beauty of JEE mathematics—the way complex, messy expressions collapse into simple, elegant truths. You have successfully navigated the intersection.

Animated Solution for Mathematics - Differentiation: An angle of intersection of the curves, $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and $x^{2} + y^{2} = ab, a > b$ , is :

Select Answer:

Visualized Solution

Visualizing the Intersection

Slope of Ellipse Tangent $m_{1}$

Slope of Circle Tangent $m_{2}$

Setting up Intersection Point

Solving for $x_{1}^{2}$

Solving for $y_{1}^{2}$

Ratio of Coordinates

The Angle Formula

Substituting Slopes

Simplifying Numerator

Simplifying Denominator

Final Calculation of $tan θ$

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus of Slopes

The Algebraic Dance

The Final Convergence

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An angle of intersection of the curves, a2x2​+b2y2​=1 and x2+y2=ab,a>b, is :

Select Answer:

Visualized Solution

Visualizing the Intersection

Slope of Ellipse Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​

Slope of Circle Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​