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given the equation, $\frac{(x - 3)^2}{49}+\frac{(y - 9)^2}{4}=1$, what …

Question

given the equation, $\frac{(x - 3)^2}{49}+\frac{(y - 9)^2}{4}=1$, what is the length of the major axis?
2 units
14 units
49 units
7 units

Explanation:

Step1: Identify the form of ellipse equation

The standard - form of an ellipse is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$. In the given equation $\frac{(x - 3)^2}{49}+\frac{(y - 9)^2}{4}=1$, we have $a^2 = 49$ and $b^2=4$.

Step2: Determine the value of a

Since $a^2 = 49$, then $a=\sqrt{49}=7$ (we take the positive value as $a$ represents a distance).

Step3: Recall the formula for the length of the major - axis

For an ellipse $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$, if $a>b$, the length of the major - axis is $2a$.

Step4: Calculate the length of the major - axis

Substitute $a = 7$ into $2a$, we get $2\times7 = 14$.

Answer:

B. 14 units