The locus of mid-points of the line segments joining (-3, -5) and the points on the ellipse x²/4 + y

Q: Why do we use parametric coordinates $(2 \cos \theta, 3 \sin \theta)$ instead of just $(x, y)$?

Great question! While you could use $(x, y)$, you would be stuck with the constraint $\frac{x^2}{4} + \frac{y^2}{9} = 1$ throughout your algebra. By using the parameter $\theta$, we transform the geometric constraint into a simple trigonometric identity, $\cos^2 \theta + \sin^2 \theta = 1$, which makes the algebra much cleaner and less prone to errors.

Q: Why do we replace $(h, k)$ with $(x, y)$ at the end?

The variables $(h, k)$ represent the coordinates of a specific midpoint on the locus. To define the entire path (the locus itself), we need a general equation in terms of $x$ and $y$. Replacing $(h, k)$ with $(x, y)$ is the standard way to transition from a specific point to the general equation of the curve.

Q: Is the resulting locus always an ellipse?

Yes! In this case, because we are essentially scaling and shifting the original ellipse, the resulting locus is also an ellipse. The coefficients of $x^2$ and $y^2$ remain positive and unequal, which confirms it is an ellipse.

Question

The locus of mid-points of the line segments joining $(-3, -5)$ and the points on the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is :

Options:

(A) $9x^2 + 4y^2 + 18x + 8y + 145 = 0$

(B) $36x^2 + 16y^2 + 90x + 56y + 145 = 0$

(C) $36x^2 + 16y^2 + 108x + 80y + 145 = 0$

(D) $36x^2 + 16y^2 + 72x + 32y + 145 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are working with an ellipse defined by the equation:

$$
\frac{x^2}{4} + \frac{y^2}{9} = 1
$$

We are given a fixed point $B(-3, -5)$. Our objective is to determine the locus of the midpoint $M(h, k)$ of the segment $BP$, where $P$ is any point lying on the ellipse.

### The Parametric Dance

To simplify the geometry, we represent any point $P$ on the ellipse using parametric coordinates. We define $P$ as:

$$
P = (2 \cos 	heta, 3 \sin 	heta)
$$

By using the parameter $	heta$, we transform the geometric constraint of the ellipse into a manageable trigonometric form. This allows us to describe every point on the ellipse using a single variable.

### The Midpoint Bridge

The midpoint $M(h, k)$ of the segment $BP$ is the average of the coordinates of $B(-3, -5)$ and $P(2 \cos 	heta, 3 \sin 	heta)$. We can express the coordinates of $M$ as:

$$
h = \frac{2 \cos 	heta - 3}{2}, \quad k = \frac{3 \sin 	heta - 5}{2}
$$

These equations serve as the bridge between the moving point $P$ and our target midpoint $M$.

### The Algebraic Alchemy

To find the locus, we must eliminate the parameter $	heta$. First, we isolate $\cos 	heta$ and $\sin 	heta$ from our midpoint equations:

$$
\cos 	heta = \frac{2h + 3}{2}, \quad \sin 	heta = \frac{2k + 5}{3}
$$

We now apply the fundamental trigonometric identity $\cos^2 	heta + \sin^2 	heta = 1$. Substituting our expressions into this identity yields:

$$
\left( \frac{2h + 3}{2} ight)^2 + \left( \frac{2k + 5}{3} ight)^2 = 1
$$

### The Final Reveal

Expanding the squares, we obtain:

$$
\frac{4h^2 + 12h + 9}{4} + \frac{4k^2 + 20k + 25}{9} = 1
$$

To clear the fractions, we multiply the entire equation by $36$:

$$
9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36
$$

Distributing the constants results in:

$$
36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36
$$

Combining like terms and replacing $(h, k)$ with $(x, y)$, we arrive at the final equation of the locus:

$$
**36x^2 + 16y^2 + 108x + 80y + 145 = 0**
$$

This result confirms that the locus of the midpoint is another ellipse, scaled and shifted relative to the original.

Animated Solution for Mathematics - Conic Sections: The locus of mid-points of the line segments joining $(- 3, - 5)$ and the points on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ is :

Select Answer:

Visualized Solution

Identify the Ellipse and Fixed Point

Define Parametric Point $P$

Connect and Find Midpoint

Apply Midpoint Formula

Isolate Trigonometric Terms

The Trigonometric Identity

Substitute and Setup

Expand the Squares

Clear the Fractions

Distribute and Simplify

The Final Locus Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Parametric Dance

The Midpoint Bridge

The Algebraic Alchemy

The Final Reveal

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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 locus of mid-points of the line segments joining (−3,−5) and the points on the ellipse 4x2​+9y2​=1 is :