Mathematics: Solve the Problem. What is x?

Solve the Problem. What is x? - id5858606, rakanmadi87

Ответ:

ldibut2003

08.07.2021

Mathematics

correct answer is option 3:

$\dfrac{x^{2} }{30}+\dfrac{y^{2} }{5}=1$

Step-by-step explanation:

Given that

Ellipse passes through coordinates:

$\approx (5\dfrac{1}{2},0),\approx (-5\dfrac{1}{2},0), \approx (0,2)\ and\ \approx (0,-2)$

Points inside ellipse (5, 0) and (-5,0).

Given the options:

$1.\ \dfrac{x^{2} }{5}+\dfrac{y^{2} }{30}=1\\2.\ \dfrac{x^{2} }{5}-\dfrac{y^{2} }{30}=1\\3.\ \dfrac{x^{2} }{30}+\dfrac{y^{2} }{5}=1\\4.\ \dfrac{x^{2} }{30}+\dfrac{y^{2} }{25}=1\\$

Let us try to put the coordinates in the options and check which equation gets satisfied.

Putting x = 0, and trying to find out y in option 1.

$\ \dfrac{0^{2} }{5}+\dfrac{y^{2} }{30}=1\\\Rightarrow \dfrac{y^{2} }{30}=1\\\Rightarrow y =\pm\sqrt{30} \approx \pm5\dfrac{1}{2}$

But, the points are $\approx (5\dfrac{1}{2},0),\approx (-5\dfrac{1}{2},0), \approx (0,2)\ and\ \approx (0,-2)$

So, option (1) is false.

Putting x = 0 to find y in option 2.

$\ \dfrac{0^{2} }{5}-\dfrac{y^{2} }{30}=1\\\Rightarrow \dfrac{y^{2} }{30}=-1$

y will have imaginary values so ellipse not possible at x = 0.

So, option (2) is false.

Putting x = 0 to find y in option 3.

$\ \dfrac{0^{2} }{30}+\dfrac{y^{2} }{5}=1\\\Rightarrow \dfrac{y^{2} }{5}=1\\\Rightarrow y \approx \pm2$

Putting y = 0 to find y in option 3.

$\ \dfrac{x^{2} }{30}+\dfrac{0^{2} }{5}=1\\\Rightarrow \dfrac{x^{2} }{30}=1\\\Rightarrow x \approx \pm5\dfrac{1}{2}$

So, the points $\approx (5\dfrac{1}{2},0),\approx (-5\dfrac{1}{2},0), \approx (0,2)\ and\ \approx (0,-2)$ are satisfied.

Putting (5, 0) and (-5,0) in LHS of option 3:

$\ \dfrac{(\pm5)^{2} }{30}+\dfrac{0 }{5}\\\Rightarrow \dfrac{25} {30}\\\Rightarrow \dfrac{5}{6}$

i.e. both the points are inside ellipse.

Hence correct answer is option 3:

$\dfrac{x^{2} }{30}+\dfrac{y^{2} }{5}=1$

Which of the following is the equation for the graph shown?

graph of an ellipse on a coordinate pla

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