Question

How do you find the domain and range of $ 4{x^2} + 25{y^2} = 100 $ algebraically?

Answer 1

Hint: We know that this is the equation of ellipse. To find its domain, we will first make this equation as y is a function of x. After that, we will solve it for real value which will be our domain. And for the range, we will put the value of x as zero and obtain the values for y which will be our range.

Complete step-by-step answer:
We are given the equation: $ 4{x^2} + 25{y^2} = 100 $
To find the domain we will make y as a function of x.
 $
  4{x^2} + 25{y^2} = 100 \\
   \Rightarrow 25{y^2} = 100 - 4{x^2} \\
   \Rightarrow {y^2} = \dfrac{{100 - 4{x^2}}}{{25}} \\
   \Rightarrow y = \pm \dfrac{{\sqrt {100 - 4{x^2}} }}{5} \;
  $
We know that the number under the square root must be greater than or equal to to get the real values as a solution.
 $
   \Rightarrow 100 - 4{x^2} \geqslant 0 \\
   \Rightarrow 4{x^2} \leqslant 100 \\
   \Rightarrow {x^2} \leqslant 25 \;
  $
 $ \Rightarrow x \leqslant 5 $ and $ x \geqslant - 5 $
Therefore, we can say that the domain of the given equation is $ \left[ { - 5,5} \right] $ .
Now, to find the range, we will put the value of x as zero in the given equation.
 $
  4{x^2} + 25{y^2} = 100 \\
   \Rightarrow 4(0) + 25{y^2} = 100 \\
   \Rightarrow 25{y^2} = 100 \\
   \Rightarrow {y^2} = 4 \\
   \Rightarrow y = \pm 2 \;
  $
Thus, the range of the given equation is $ \left[ { - 2,2} \right] $ .
This, by this method we can determine that the domain of $ 4{x^2} + 25{y^2} = 100 $ is $ \left[ { - 5,5} \right] $ and its range is $ \left[ { - 2,2} \right] $ .

Note: Here, we have solved this question algebraically. However, if we simply observe the equation, we can say that $ {x^2} $ cannot be greater than 25 as the least possible value for $ {y^2} $ is zero and thus the domain is $ \left[ { - 5,5} \right] $ . Similarly, $ {y^2} $ cannot be greater than 4 as the least possible value for $ {x^2} $ is zero.