Chord with Given Midpoint

The equation of a chord of an ellipse with a known midpoint $$(x_1, y_1)$$ is given by the formula:

$$T = S_1$$

Step 1: Set up $$T$$ and $$S_1$$

The given ellipse is:

$$\frac{x^2}{16} + \frac{y^2}{9} - 1 = 0$$

The given midpoint is $$(x_1, y_1) = \left(2, \frac{1}{2}\right)$$.

• To find $$T$$ (replace $$x^2 \to x x_1$$ and $$y^2 \to y y_1$$):$$T = \frac{x(2)}{16} + \frac{y\left(\frac{1}{2}\right)}{9} - 1 = \frac{x}{8} + \frac{y}{18} - 1$$
• To find $$S_1$$ (substitute the midpoint into the ellipse equation):$$S_1 = \frac{(2)^2}{16} + \frac{\left(\frac{1}{2}\right)^2}{9} - 1 = \frac{4}{16} + \frac{\frac{1}{4}}{9} - 1 = \frac{1}{4} + \frac{1}{36} - 1$$

Step 2: Equate $$T = S_1$$

Cancel out the $$-1$$ from both sides:

$$\frac{x}{8} + \frac{y}{18} = \frac{1}{4} + \frac{1}{36}$$

Simplify the right side by finding a common denominator ($$36$$):

$$\frac{1 \cdot 9}{36} + \frac{1}{36} = \frac{10}{36} = \frac{5}{18}$$

Now update the equation:

$$\frac{x}{8} + \frac{y}{18} = \frac{5}{18}$$

Step 3: Align with the Target Form

The required form is $$\alpha x + \beta y = 41$$. To clear the fractions and see how to get $$41$$ on the right side, let's first find the least common multiple (LCM) of the denominators on the left side ($$8$$ and $$18$$), which is $$72$$.

Multiply the entire equation by $$72$$:

$$72 \left( \frac{x}{8} + \frac{y}{18} \right) = 72 \left( \frac{5}{18} \right)$$ $$9x + 4y = 4 \cdot 5$$ $$9x + 4y = 20$$

We need the right side to be $$41$$ instead of $$20$$. To adjust this, multiply the entire equation by $$\frac{41}{20}$$:

$$\frac{41}{20}(9x) + \frac{41}{20}(4y) = 41$$  $$\frac{369}{20}x + \frac{41}{5}y = 41$$

Comparing this with $$\alpha x + \beta y = 41$$, we find:

• $$\alpha = \frac{369}{20}$$
• $$\beta = \frac{41}{5} = \frac{164}{20}$$

Final Calculation

$$\alpha + \beta = \frac{369}{20} + \frac{164}{20} = \frac{533}{20} = 26.65$$ 

Let the midpoint of the chord be $$P(h, k)$$.

The equation of a chord of the ellipse whose midpoint is given as $$(h, k)$$ can be written using the standard formula $$T = S_1$$:

$$\frac{hx}{16} + \frac{ky}{9} = \frac{h^2}{16} + \frac{k^2}{9}$$

Let $$C = \frac{h^2}{16} + \frac{k^2}{9}$$. We can rewrite the equation of the chord to express it as equal to 1:

$$\frac{1}{C}\left(\frac{hx}{16} + \frac{ky}{9}\right) = 1 \quad \cdots(1)$$

The equation of the given ellipse is:

$$\frac{x^2}{16} + \frac{y^2}{9} = 1 \quad \cdots(2)$$

To find the lines connecting the center (origin) to the intersection points of the chord and the ellipse, we homogenize equation (2) using equation (1):

$$\frac{x^2}{16} + \frac{y^2}{9} = \left[\frac{1}{C}\left(\frac{hx}{16} + \frac{ky}{9}\right)\right]^2$$

$$\frac{x^2}{16} + \frac{y^2}{9} = \frac{1}{C^2}\left(\frac{h^2x^2}{256} + \frac{k^2y^2}{81} + \frac{2hkxy}{144}\right)$$

Rearranging the terms into a quadratic form in terms of $$x$$ and $$y$$:

$$\left(\frac{1}{16} - \frac{h^2}{256C^2}\right)x^2 - \frac{2hk}{144C^2}xy + \left(\frac{1}{9} - \frac{k^2}{81C^2}\right)y^2 = 0$$

Since the chord subtends a right angle at the center, the pair of straight lines represented by this homogenized equation must be perpendicular to each other. Therefore, the sum of the coefficients of $$x^2$$ and $$y^2$$ must be zero:

$$\left(\frac{1}{16} - \frac{h^2}{256C^2}\right) + \left(\frac{1}{9} - \frac{k^2}{81C^2}\right) = 0$$

$$\frac{1}{16} + \frac{1}{9} = \frac{1}{C^2}\left(\frac{h^2}{256} + \frac{k^2}{81}\right)$$

$$\frac{25}{144} = \frac{1}{C^2}\left(\frac{h^2}{256} + \frac{k^2}{81}\right)$$

Multiplying both sides by $$144C^2$$:

$$25C^2 = 144\left(\frac{h^2}{256} + \frac{k^2}{81}\right)$$

$$25C^2 = \frac{144h^2}{256} + \frac{144k^2}{81}$$

Simplifying the coefficients on the right-hand side ($$\frac{144}{256} = \frac{9}{16}$$ and $$\frac{144}{81} = \frac{16}{9}$$):

$$25C^2 = \frac{9h^2}{16} + \frac{16k^2}{9}$$

Replacing $$C$$ back with its definition $$\frac{h^2}{16} + \frac{k^2}{9}$$:

$$25\left(\frac{h^2}{16} + \frac{k^2}{9}\right)^2 = \frac{9h^2}{16} + \frac{16k^2}{9}$$

Replacing $$(h, k)$$ with $$(x, y)$$ to get the final locus:

$$25\left(\frac{x^2}{16} + \frac{y^2}{9}\right)^2 = \frac{9x^2}{16} + \frac{16y^2}{9}$$