Question

How do you find the centre, vertices, foci and eccentricity of \[9{{x}^{2}}+25{{y}^{2}}=225\]?

Answer 1

Hint: This question is from the topic of ellipse. In this question, we will first divide the term 225 to the both sides of the equation. After that, we will understand how to find centre, vertices, foci and eccentricity of ellipse from the general equation of ellipse. After that, we will know the general equation of ellipse. After that, we will find the centre, vertices, foci and eccentricity using the equation \[9{{x}^{2}}+25{{y}^{2}}=225\].

Complete step by step solution:
Let us solve this question.
In this question, it is asked to find the vertices, centre, foci and eccentricity of the equation of ellipse which is given in the question as \[9{{x}^{2}}+25{{y}^{2}}=225\]
After dividing 225 to the both side of equation, we can write the equation \[9{{x}^{2}}+25{{y}^{2}}=225\] as
\[\Rightarrow \dfrac{9}{225}{{x}^{2}}+\dfrac{25}{225}{{y}^{2}}=\dfrac{225}{225}\]
Now, let us understand the formula of the general equation of ellipse.
The equation is \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\] general equation of ellipse.
From the above equation of ellipse, we can say that
The centre of ellipse will be \[(h,k)\]
Foci of ellipse will be \[(h+c,k)\text{ and }(h-c,k)\]
Vertices of ellipse will be \[(h+a,k)\text{ and }(h-a,k)\]
Eccentricity of ellipse will be \[e=\dfrac{c}{a}=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}\]
Now, we have equation of ellipse as
\[\dfrac{9}{225}{{x}^{2}}+\dfrac{25}{225}{{y}^{2}}=\dfrac{225}{225}\]
As we know that, if we divide 225 by 9 then we get 25 and if we divide 225 by 25 then we get 9, so we can write the above equation as
\[\Rightarrow \dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1\]
The above equation can also be written as
\[\Rightarrow \dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\]
The above equation can also be written as
\[\Rightarrow \dfrac{{{\left( x-0 \right)}^{2}}}{{{5}^{2}}}+\dfrac{{{\left( y-0 \right)}^{2}}}{{{3}^{2}}}=1\]
From the above ellipse equation, we can say that
a=5, b=3, h=0, and k=0
\[c=\sqrt{{{a}^{2}}-{{b}^{2}}}=\sqrt{{{5}^{2}}-{{3}^{2}}}=\sqrt{25-9}=\sqrt{16}=4\]
Now, we can see that
The centre is (0,0)
Foci are (4,0) and (-4,0)
Vertices are (5,0) and (-5,0)
Eccentricity is \[e=\dfrac{c}{a}=\dfrac{4}{5}\]
Now, we have found the centre, eccentricity, vertices and foci.
We can take reference from the following figure.

Note: We should have a better knowledge in the topic of ellipse which is from the chapter coordinate geometry.
Remember that the general equation of ellipse is \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\]
Always remember the square of some specific numbers like the square of 15 is 225, the square 0f 3 is 9 and the square of 5 is 25.
Remember of formula of eccentricity that is \[e=\dfrac{c}{b}=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\]