The acute angle between the pair of tangents drawn to the ellipse 2x² + 3y² = 5 from the point (1,3)

Q: Wait, why do we square both sides? Doesn't that introduce extraneous roots?

That is a brilliant observation! Yes, squaring can introduce extraneous roots. However, in this specific context, we are looking for the slopes of the two tangents. The quadratic equation we derive, $9m^2 + 36m - 44 = 0$, correctly captures the slopes of both tangents. Since we only need the sum and product of these slopes to find the angle, the quadratic approach is perfectly valid and efficient.

Q: Why can't we just use the point-of-contact form for the tangent?

You absolutely could! You could assume the point of contact is $(x_1, y_1)$ and use the equation $2xx_1 + 3yy_1 = 5$. Since it passes through $(1, 3)$, you get $2x_1 + 9y_1 = 5$. Combined with the fact that $(x_1, y_1)$ lies on the ellipse, you could solve for the points of contact. However, the slope form is often faster for finding the angle between tangents because it directly gives you a quadratic in $m$.

Q: How do I know if the angle is acute or obtuse?

The formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|$ gives the tangent of the acute angle because of the absolute value. If the result is positive, it's the acute angle. If you were asked for the obtuse angle, you would simply take $180^\circ - \theta$.

Question

The acute angle between the pair of tangents drawn to the ellipse $2x^2 + 3y^2 = 5$ from the point $(1,3)$ is

Options:

(A) $	an^{-1}\left(\frac{16}{7\sqrt{5}}ight)$

(B) $	an^{-1}\left(\frac{24}{7\sqrt{5}}ight)$

(C) $	an^{-1}\left(\frac{32}{7\sqrt{5}}ight)$

(D) $	an^{-1}\left(\frac{3+8\sqrt{5}}{35}ight)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe! Today, we are going to unravel a classic problem in coordinate geometry. We are standing before an ellipse, defined by the equation $2x^2 + 3y^2 = 5$, and we are tasked with finding the acute angle between two tangents drawn from an external point $P(1, 3)$.

This is not just a calculation; it is a story of how we translate geometric intuition into the language of algebra.

### Standardizing the Vision

Before we can dance with the tangents, we must see the ellipse in its true form. The equation $2x^2 + 3y^2 = 5$ is currently disguised.

To bring it into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we divide the entire equation by $5$:

$$\frac{x^2}{5/2} + \frac{y^2}{5/3} = 1$$

Now, the parameters are clear: $a^2 = 5/2$ and $b^2 = 5/3$. This is our foundation.

### The Power of the Slope Form

How do we represent a tangent line? We could use the point-of-contact form, but that leads us into a labyrinth of coordinates. Instead, let us use the slope form.

Any line with slope $m$ that is tangent to our ellipse must satisfy the condition:

$$y = mx \pm \sqrt{a^2m^2 + b^2}$$

Substituting our values, we get:

$$y = mx \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$$

This equation is our key. It represents every possible tangent to this ellipse.

### The Quadratic Trap

We know our tangent must pass through the point $P(1, 3)$. This is a constraint that forces the line to be one of the two specific tangents.

Substituting $x=1$ and $y=3$ into our equation, we get:

$$3 - m = \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$$

To solve for $m$, we square both sides:

$$(3 - m)^2 = \frac{5}{2}m^2 + \frac{5}{3}$$

Expanding this, we obtain $9 - 6m + m^2 = \frac{5}{2}m^2 + \frac{5}{3}$. Multiplying by $6$ to clear the denominators and rearranging terms, we arrive at the quadratic equation:

$$9m^2 + 36m - 44 = 0$$

This quadratic is the heart of our problem. Its roots, $m_1$ and $m_2$, are the slopes of our two tangents.

### The Final Flourish

We do not need to solve for $m_1$ and $m_2$ individually. We only need their sum and product. From the quadratic, $m_1 + m_2 = -4$ and $m_1m_2 = -44/9$.

The angle $	heta$ between the lines is given by:

$$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$$

Using the identity $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2}$, we calculate the numerator:

$$\sqrt{(-4)^2 - 4\left(-\frac{44}{9}ight)} = \sqrt{16 + \frac{176}{9}} = \sqrt{\frac{320}{9}} = \frac{8\sqrt{5}}{3}$$

The denominator is:

$$\left| 1 + m_1m_2 ight| = \left| 1 - \frac{44}{9} ight| = \left| -\frac{35}{9} ight| = \frac{35}{9}$$

Finally, calculating the tangent:

$$	an 	heta = \frac{8\sqrt{5}/3}{35/9} = \frac{24\sqrt{5}}{35}$$

Thus, the acute angle is **$	heta = 	an^{-1}\left(\frac{24\sqrt{5}}{35}ight)$**. We have conquered the problem, not by brute force, but by the elegance of algebraic structure.

Animated Solution for Mathematics - Conic Sections: The acute angle between the pair of tangents drawn to the ellipse $2 x^{2} + 3 y^{2} = 5$ from the point $(1, 3)$ is

Select Answer:

Visualized Solution

Visualizing the Ellipse and the Point

Standardizing the Ellipse Equation

The Slope Form of the Tangent

Passing through Point $(1, 3)$

Squaring to Remove the Radical

Forming the Quadratic Equation

Sum and Product of Slopes

The Angle Formula

Calculating the Numerator

Calculating the Denominator

Final Calculation of Tan Theta

Conclusion and Key Takeaways

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Vision

The Power of the Slope Form

The Quadratic Trap

The Final Flourish

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=5 from the point (1,3) is