
How do you find the center and vertices of the ellipse \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1\]?
Answer
549.3k+ views
Hint: The standard form of equation of ellipse is \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. And a, b are any real numbers. We can find the center and vertices using the equation and the values of a and b. To do this, first, we need to express the given equation in its standard form. The centre of the ellipse with the equation \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[x=0\And y=0\]. The vertices are similar to the intercepts of the ellipse.
Complete step by step solution:
We are given the equation of the ellipse as \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1\]. Comparing it with the standard form of the equation \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we get \[{{a}^{2}}=4\And {{b}^{2}}=\dfrac{1}{4}\]. Taking the square root of both these equations to find the values, we get \[a=2\And b=\dfrac{1}{2}\].
We know that to find the center of the ellipse, we have to equate the x and y with zero. Thus, we get coordinates of centre as \[(0,0)\].
The vertices are the intercepts of the ellipse; thus, we can find them by substituting x and y to be zero separately.
Substituting \[x=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{{{0}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1 \\
& \Rightarrow \dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1 \\
\end{align}\]
Solving the above equation, we get \[y=\pm \dfrac{1}{2}\]. Thus, the coordinates of the two vertices are \[\left( 0,\dfrac{1}{2} \right)\And \left( 0,-\dfrac{1}{2} \right)\].
Substituting \[y=0\],
\[\begin{align}
& \Rightarrow \dfrac{{{x}^{2}}}{4}+\dfrac{{{0}^{2}}}{\dfrac{1}{4}}=1 \\
& \Rightarrow \dfrac{{{x}^{2}}}{4}=1 \\
\end{align}\]
Solving the above equation, we get \[x=\pm 2\]. Thus, the coordinates of the other two vertices are \[\left( 2,0 \right)\And \left( -2,0 \right)\].
We can use the equation to graph the ellipse as,
Note: To solve these types of questions based on the conics. One should know the different properties of the conics as well as how to find the components like center, vertex, etc. using the equation of the conic. The other conics whose properties we should remember are straight lines, circle, hyperbola, parabola.
Complete step by step solution:
We are given the equation of the ellipse as \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1\]. Comparing it with the standard form of the equation \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we get \[{{a}^{2}}=4\And {{b}^{2}}=\dfrac{1}{4}\]. Taking the square root of both these equations to find the values, we get \[a=2\And b=\dfrac{1}{2}\].
We know that to find the center of the ellipse, we have to equate the x and y with zero. Thus, we get coordinates of centre as \[(0,0)\].
The vertices are the intercepts of the ellipse; thus, we can find them by substituting x and y to be zero separately.
Substituting \[x=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{{{0}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1 \\
& \Rightarrow \dfrac{{{y}^{2}}}{\dfrac{1}{4}}=1 \\
\end{align}\]
Solving the above equation, we get \[y=\pm \dfrac{1}{2}\]. Thus, the coordinates of the two vertices are \[\left( 0,\dfrac{1}{2} \right)\And \left( 0,-\dfrac{1}{2} \right)\].
Substituting \[y=0\],
\[\begin{align}
& \Rightarrow \dfrac{{{x}^{2}}}{4}+\dfrac{{{0}^{2}}}{\dfrac{1}{4}}=1 \\
& \Rightarrow \dfrac{{{x}^{2}}}{4}=1 \\
\end{align}\]
Solving the above equation, we get \[x=\pm 2\]. Thus, the coordinates of the other two vertices are \[\left( 2,0 \right)\And \left( -2,0 \right)\].
We can use the equation to graph the ellipse as,
Note: To solve these types of questions based on the conics. One should know the different properties of the conics as well as how to find the components like center, vertex, etc. using the equation of the conic. The other conics whose properties we should remember are straight lines, circle, hyperbola, parabola.
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