Find the set of those value(s) of `alpha` for which `(7-(5alpha)/4,alpha)` lies inside the ellipse `x^2/25+y^2/16=1`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the set of values of \( \alpha \) for which the point \( \left( 7 - \frac{5\alpha}{4}, \alpha \right) \) lies inside the ellipse defined by the equation \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \), we need to follow these steps:

### Step 1: Substitute the point into the ellipse equation
The point \( \left( 7 - \frac{5\alpha}{4}, \alpha \right) \) must satisfy the inequality derived from the ellipse equation:

\[ \frac{\left( 7 - \frac{5\alpha}{4} \right)^2}{25} + \frac{\alpha^2}{16} < 1 \]

### Step 2: Simplify the inequality
First, we simplify the left-hand side:

1. Calculate \( \left( 7 - \frac{5\alpha}{4} \right)^2 \):
\[ \left( 7 - \frac{5\alpha}{4} \right)^2 = 49 - 2 \cdot 7 \cdot \frac{5\alpha}{4} + \left( \frac{5\alpha}{4} \right)^2 = 49 - \frac{70\alpha}{4} + \frac{25\alpha^2}{16} \]
\[ = 49 - \frac{35\alpha}{2} + \frac{25\alpha^2}{16} \]

2. Substitute this back into the inequality:
\[ \frac{49 - \frac{35\alpha}{2} + \frac{25\alpha^2}{16}}{25} + \frac{\alpha^2}{16} < 1 \]

### Step 3: Clear the denominators
Multiply through by 400 (the least common multiple of 25 and 16) to eliminate the fractions:
\[ 16 \left( 49 - \frac{35\alpha}{2} + \frac{25\alpha^2}{16} \right) + 25\alpha^2 < 400 \]

### Step 4: Expand and combine like terms
Expanding gives:
\[ 784 - 280\alpha + 25\alpha^2 + 25\alpha^2 < 400 \]
\[ 50\alpha^2 - 280\alpha + 784 - 400 < 0 \]
\[ 50\alpha^2 - 280\alpha + 384 < 0 \]

### Step 5: Factor the quadratic inequality
We can factor the quadratic:
\[ 50\alpha^2 - 280\alpha + 384 = 0 \]
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \alpha = \frac{280 \pm \sqrt{(-280)^2 - 4 \cdot 50 \cdot 384}}{2 \cdot 50} \]
Calculating the discriminant:
\[ 78400 - 76800 = 1600 \]
Thus:
\[ \alpha = \frac{280 \pm 40}{100} \]
Calculating the roots:
\[ \alpha_1 = \frac{320}{100} = 3.2, \quad \alpha_2 = \frac{240}{100} = 2.4 \]

### Step 6: Determine the intervals
The quadratic opens upwards (since the coefficient of \( \alpha^2 \) is positive), so the inequality \( 50\alpha^2 - 280\alpha + 384 < 0 \) holds between the roots:
\[ 2.4 < \alpha < 3.2 \]

### Final Answer:
The set of values of \( \alpha \) for which the point \( \left( 7 - \frac{5\alpha}{4}, \alpha \right) \) lies inside the ellipse is:
\[ \alpha \in (2.4, 3.2) \]