Let θ be the acute angle between the tangents to the ellipse x²/9 + y²/1 = 1 and the circle x² + y²

Q: Wait, why do we need to find the intersection point first? Can't we just find the angle generally?

Great question! The angle between two curves is not constant; it changes depending on where they intersect. Because these curves intersect at specific points, the slopes of their tangents are unique to that location. We must find the specific coordinates of the intersection point $P$ to determine the exact slopes $m_1$ and $m_2$ at that precise moment.

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

The formula $\frac{m_1 - m_2}{1 + m_1 m_2}$ can yield a negative value depending on which slope you subtract from the other. However, the problem specifically asks for the *acute* angle $\theta$. Since the tangent of an acute angle must be positive, the absolute value ensures our result is positive and geometrically correct.

Q: Is there a faster way to find the slopes without doing full implicit differentiation?

For these standard conic sections, you could memorize the tangent equations, but implicit differentiation is the most robust and 'JEE-proof' method. It works for any curve, not just conics. Mastering this technique ensures you won't get stuck if the problem involves more complex, non-standard curves in the future.

Question

Let $	heta$ be the acute angle between the tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{1} = 1$ and the circle $x^2 + y^2 = 3$ at their point of intersection in the first quadrant. Then $	an 	heta$ is equal to :

Options:

(A) $\frac{5}{2\sqrt{3}}$

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

(C) $\frac{4}{\sqrt{3}}$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the acute angle $	heta$ between the tangents of the ellipse $\frac{x^2}{9} + y^2 = 1$ and the circle $x^2 + y^2 = 3$ at their point of intersection in the first quadrant.

### The Hunt for the Intersection Point

To find the point of intersection, we treat the two equations as a system. From the circle equation, we have $y^2 = 3 - x^2$. Substituting this into the ellipse equation:

$$\frac{x^2}{9} + (3 - x^2) = 1$$

Multiplying the entire equation by $9$ to clear the denominator yields:

$$x^2 + 27 - 9x^2 = 9$$

Simplifying this expression results in $-8x^2 = -18$, which leads to $x^2 = \frac{18}{8} = \frac{9}{4}$. Since we are restricted to the first quadrant, we find $x = \frac{3}{2}$.

Plugging this value back into the circle equation gives $y^2 = 3 - \frac{9}{4} = \frac{3}{4}$, so $y = \frac{\sqrt{3}}{2}$. Thus, the intersection point is $P\left(\frac{3}{2}, \frac{\sqrt{3}}{2}ight)$.

### The Art of Differentiation

To find the slopes of the tangents, we use implicit differentiation. For the ellipse $\frac{x^2}{9} + y^2 = 1$:

$$\frac{2x}{9} + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{9y}$$

Substituting the coordinates of point $P$, the slope $m_1$ is:

$$m_1 = -\frac{3/2}{9(\sqrt{3}/2)} = -\frac{1}{3\sqrt{3}}$$

For the circle $x^2 + y^2 = 3$, differentiating gives $2x + 2y \frac{dy}{dx} = 0$, which simplifies to $\frac{dy}{dx} = -\frac{x}{y}$. Substituting point $P$ again, the slope $m_2$ is:

$$m_2 = -\frac{3/2}{\sqrt{3}/2} = -\sqrt{3}$$

### The Geometric Synthesis

The angle $	heta$ between these two lines is determined by the formula:

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

Substituting our calculated slopes $m_1 = -\frac{1}{3\sqrt{3}}$ and $m_2 = -\sqrt{3}$:

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

The numerator simplifies to $\sqrt{3} - \frac{1}{3\sqrt{3}} = \frac{9-1}{3\sqrt{3}} = \frac{8}{3\sqrt{3}}$. The denominator simplifies to $1 + \frac{1}{3} = \frac{4}{3}$.

Performing the final division:

$$	an 	heta = \frac{8}{3\sqrt{3}} 	imes \frac{3}{4} = \frac{2}{\sqrt{3}}$$

The acute angle between the tangents is given by **$	heta = \arctan\left(\frac{2}{\sqrt{3}}ight)$**.

Animated Solution for Mathematics - Differentiation: Let $θ$ be the acute angle between the tangents to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1$ and the circle $x^{2} + y^{2} = 3$ at their point of intersection in the first quadrant. Then $tan θ$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curves

Finding the Intersection Point

Substituting $y^{2}$

Solving for $x^{2}$

Calculating Coordinates of $P$

Slope of the Ellipse Tangent ( $m_{1}$ )

Calculating $m_{1}$ at Point $P$

Slope of the Circle Tangent ( $m_{2}$ )

Calculating $m_{2}$ at Point $P$

The Angle Formula

Substituting the Slopes

Final Computation

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Hunt for the Intersection Point

The Art of Differentiation

The Geometric Synthesis

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Select Answer:

Visualized Solution

Visualizing the Curves

Finding the Intersection Point

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

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

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​ at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​ at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P