Question

How do you find centre, vertex and foci of an ellipse $9{{x}^{2}}+25{{y}^{2}}-36x+50y-164=0$?

Answer 1

In this problem, we have to find centre, vertex and foci of an ellipse $9{{x}^{2}}+25{{y}^{2}}-36x+50y-164=0$. We can first draw an ellipse. We can then analyse the equation we can say the given equation is not as the general form of an ellipse. We can convert the given equation into an equation of ellipse form by taking common terms from the equation and dividing it. After finding the equation of ellipse we can find the centre, vertex and foci by using the formulas.

Complete step by step solution:
We know that the given equation of an ellipse is,
$9{{x}^{2}}+25{{y}^{2}}-36x+50y-164=0$…….. (1)
We also know that the general form of an ellipse is,
$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1,a>b$ ……… (2)
Where, $\left( h,k \right)$ is the centre, $\left( \pm a,0 \right)$ is the vertex, $\left( \pm ae,0 \right)$ is the foci.
We can take common terms from the equation (1) then we can convert the step into a perfect square to get a whole square form and add the same terms to the right-hand side, we get

& \Rightarrow 9{{x}^{2}}-36x+25{{y}^{2}}+50y=164 \\\ & \Rightarrow 9\left( {{x}^{2}}-4x \right)+25\left( {{y}^{2}}+2y \right)=164 \\\ & \Rightarrow 9\left( {{x}^{2}}-4x+4 \right)+25\left( {{y}^{2}}+2y+1 \right)=164+36+25 \\\ & \Rightarrow 9{{\left( x-2 \right)}^{2}}+25{{\left( y+1 \right)}^{2}}=225 \\\ \end{aligned}$$ We can now divide 225 on both sides, we get $$\Rightarrow \dfrac{{{\left( x-2 \right)}^{2}}}{{{5}^{2}}}+\dfrac{{{\left( y+1 \right)}^{2}}}{{{3}^{2}}}=1$$ ……… (3) Now we can compare the equation (2) and (3), we get a = 5, b = 3 and $$\left( h,k \right)=\left( 2,-1 \right)$$. The centre of the ellipse is $$\left( 2,-1 \right)$$. The vertices are $$\left( a,0 \right),\left( -a,0 \right)$$. $$\begin{aligned} & \Rightarrow x-2=5,x-2=-5 \\\ & \Rightarrow x=7,-3 \\\ & \Rightarrow y+1=0 \\\ & \Rightarrow y=-1 \\\ \end{aligned}$$ Therefore, the vertices are $$\left( 7,-1 \right),\left( -3,-1 \right)$$. The major axis is 2a = 10. $$\Rightarrow {{e}^{2}}={{a}^{2}}-{{b}^{2}}=25-9=16={{4}^{2}}$$ From the above step, we can say that the foci are 4 units on the major axis on either side of centre. The foci are $$\left( -2,-1 \right),\left( 6,-1 \right)$$. Therefore, the centre of the ellipse is $$\left( 2,-1 \right)$$, the vertices are $$V1\left( -3,-1 \right),V2\left( 7,-1 \right)$$, the foci are $$f1\left( -2,-1 \right),f2\left( 6,-1 \right)$$.  **Note:** Students make mistakes while finding the focus of the given ellipse. We should remember that foci are 4 units on the major axis on either side of centre. We should also know how to find whether the ellipse is horizontal or vertical by seeing the equation, if the major axis of the ellipse is along the y-axis then it is vertical, if the major axis is along the x axis, then it is horizontal.