Let a circle of radius 4 be concentric to the ellipse 15x² + 19y² = 285. Then the common tangents ar

Q: Why is the minor axis the y-axis?

In the ellipse $\frac{x^2}{19} + \frac{y^2}{15} = 1$, the denominator under $x^2$ is larger than the denominator under $y^2$. This means the major axis is along the x-axis, and the minor axis is along the y-axis.

Q: Why do we use the distance formula for the circle?

A line is tangent to a circle if and only if the distance from the center to the line equals the radius. It is the most elegant way to handle tangency for circles.

Q: Why did we get $\pm$ in the tangent equation?

The $\pm$ represents the two possible tangents that are symmetric with respect to the axes. Since we are looking for the angle, the slope $m$ is what matters, and the sign just tells us which quadrant the tangent lies in.

Question

Let a circle of radius 4 be concentric to the ellipse $15x^2 + 19y^2 = 285$. Then the common tangents are inclined to the minor axis of the ellipse at the angle

Options:

(A) $\frac{\pi}{3}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the angle that the common tangent of an ellipse and a circle makes with the minor axis. Both curves are centered at the origin.

The ellipse is defined by the equation $15x^2 + 19y^2 = 285$, and the circle is defined by $x^2 + y^2 = 16$.

### Standardizing the Ellipse

To reveal the parameters of the ellipse, we divide the equation $15x^2 + 19y^2 = 285$ by $285$:

$$\frac{15x^2}{285} + \frac{19y^2}{285} = 1$$

This simplifies to the standard form:

$$\frac{x^2}{19} + \frac{y^2}{15} = 1$$

Here, we identify $a^2 = 19$ and $b^2 = 15$. Since $a^2 > b^2$, the major axis lies along the $x$-axis, and the minor axis is the $y$-axis.

### The Power of the Tangent Condition

Consider a line $y = mx + c$ that is tangent to the ellipse. The condition for tangency to an ellipse is given by:

$$c^2 = a^2m^2 + b^2$$

Substituting our known values, we obtain $c^2 = 19m^2 + 15$. Thus, any tangent to the ellipse takes the form:

$$y = mx \pm \sqrt{19m^2 + 15}$$

### The Bridge to the Circle

The circle $x^2 + y^2 = 16$ has a radius $r = 4$. For the line $mx - y \pm \sqrt{19m^2 + 15} = 0$ to be tangent to this circle, the perpendicular distance from the origin $(0,0)$ to the line must equal the radius $r$.

Using the distance formula $d = \frac{|c|}{\sqrt{m^2 + 1}}$, we set the distance equal to $4$:

$$\frac{\sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} = 4$$

Squaring both sides yields:

$$\frac{19m^2 + 15}{m^2 + 1} = 16$$

Multiplying across, we find $19m^2 + 15 = 16m^2 + 16$, which simplifies to $3m^2 = 1$, or:

$$m^2 = \frac{1}{3}$$

### Final Calculation

We have determined that $|m| = \frac{1}{\sqrt{3}}$. If $	heta$ is the angle the tangent makes with the major axis ($x$-axis), then:

$$	an 	heta = |m| = \frac{1}{\sqrt{3}}$$

This implies $	heta = 30^\circ$. Since the question asks for the angle with the **minor axis** ($y$-axis), and the axes are perpendicular, we calculate:

$$	ext{Angle} = 90^\circ - 30^\circ = 60^\circ$$

The final angle the common tangent makes with the minor axis is **$60^\circ$** (or $\frac{\pi}{3}$ radians).

Animated Solution for Mathematics - Conic Sections: Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle

Select Answer:

Visualized Solution

Visualizing the Curves

Standardizing the Ellipse Equation

Equation of the Concentric Circle

General Tangent to the Ellipse

Condition for Tangency to Circle

Raw Setup (Substitution)

Squaring the Equation

Solving for $m^{2}$

Finding the Slope $m$

Angle with the Minor Axis

Conclusion and Summary

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Ellipse

The Power of the Tangent Condition

The Bridge to the Circle

Final Calculation

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