The point on the ellipse `16x^(2)+9y^(2)=400`, where the ordinate decreases at the same rate at which the abscissa increases is (a, b), then `a+3b` can be

Options

A. 16
B. 19
C. 6
D. 9

Answer

To solve the problem, we need to find the point (a, b) on the ellipse given by the equation \(16x^2 + 9y^2 = 400\) where the ordinate (y-coordinate) decreases at the same rate as the abscissa (x-coordinate) increases.

### Step-by-Step Solution:

1. **Understanding the Rates**:
We are given that the rate at which the ordinate decreases is equal to the rate at which the abscissa increases. Mathematically, this can be expressed as:
\[ \frac{dy}{dt} = -\frac{dx}{dt} \]
Let’s denote \(\frac{dx}{dt}\) as \(r\). Therefore, we have:
\[ \frac{dy}{dt} = -r \]

2. **Differentiate the Ellipse Equation**:
We differentiate the ellipse equation \(16x^2 + 9y^2 = 400\) with respect to \(t\):
\[ \frac{d}{dt}(16x^2) + \frac{d}{dt}(9y^2) = 0 \]
Applying the chain rule, we get:
\[ 32x\frac{dx}{dt} + 18y\frac{dy}{dt} = 0 \]

3. **Substituting the Rates**:
Substitute \(\frac{dy}{dt} = -r\) into the differentiated equation:
\[ 32x \cdot r + 18y \cdot (-r) = 0 \]
Simplifying, we have:
\[ 32x r - 18y r = 0 \]
Factoring out \(r\) (assuming \(r \neq 0\)):
\[ r(32x - 18y) = 0 \]
Thus, we have:
\[ 32x = 18y \quad \Rightarrow \quad \frac{x}{y} = \frac{18}{32} = \frac{9}{16} \]
This gives us our first relation:
\[ x = \frac{9}{16}y \]

4. **Substituting into the Ellipse Equation**:
Substitute \(x = \frac{9}{16}y\) into the ellipse equation:
\[ 16\left(\frac{9}{16}y\right)^2 + 9y^2 = 400 \]
Simplifying:
\[ 16 \cdot \frac{81}{256}y^2 + 9y^2 = 400 \]
\[ \frac{1296}{256}y^2 + 9y^2 = 400 \]
Converting \(9y^2\) to a common denominator:
\[ \frac{1296}{256}y^2 + \frac{2304}{256}y^2 = 400 \]
\[ \frac{3600}{256}y^2 = 400 \]

5. **Solving for y**:
Multiply both sides by \(256\):
\[ 3600y^2 = 102400 \]
\[ y^2 = \frac{102400}{3600} = \frac{256}{9} \]
Taking the square root:
\[ y = \frac{16}{3} \]

6. **Finding x**:
Substitute \(y\) back to find \(x\):
\[ x = \frac{9}{16} \cdot \frac{16}{3} = 3 \]

7. **Calculating \(a + 3b\)**:
Now we have \(a = 3\) and \(b = \frac{16}{3}\):
\[ a + 3b = 3 + 3 \cdot \frac{16}{3} = 3 + 16 = 19 \]

### Final Answer:
Thus, the value of \(a + 3b\) is \(19\).