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If \[x=2\left( \cos \text{t}-\sin \text{t} \right)\]; \[y=3\left( \cos \text{t}+\sin \text{t} \right)\] represents a conic, its foci are:
a)\[(\pm \sqrt{10},0)\]
b)\[(\pm \sqrt{13},0)\]
c)\[(0,\pm \sqrt{13})\]
d)\[(0,\pm \sqrt{10})\]

Last updated date: 20th Jun 2024
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Hint: Firstly, try to find the conic represented by the given function by simplifying the equations using mathematical identities and get an equation in terms of x and y.
Then convert the equations into a standard form of any conic section (i.e. ellipse, parabola, or hyperbola).
Circle: $ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $
Parabola: \[\left\{ \begin{align}
  & y=\dfrac{{{\left( x-h \right)}^{2}}}{4a}+k \\
 & x=\dfrac{{{\left( y-k \right)}^{2}}}{4a}+h \\
\end{align} \right\}\]
Ellipse: \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\]
Hyperbola: \[\left\{ \begin{align}
  & \text{for a b : }\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1 \\
 & \text{for a b : }\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}-\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}=1 \\
\end{align} \right\}\]\[\left\{ \begin{align}
  & \text{for a b : }\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1 \\
 & \text{for a b : }\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}-\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}=1 \\
\end{align} \right\}\]
After getting the equation of the conic section, find the foci for the required conic section.

Complete step-by-step answer:
Consider the given functions:
  & x=2\left( \cos \text{t}-\sin \text{t} \right)......(1) \\
 & y=3\left( \cos \text{t}+\sin \text{t} \right)......(2) \\
Squaring the both sides of equation (1) & (2),
By applying the identities:\[\left[ \begin{align}
  & \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & \because {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align} \right]\]\[\left[ \begin{align}
  & \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & \because {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align} \right]\]
we get:
  & {{x}^{2}}=4\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(3) \\
 & {{y}^{2}}=9\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\

We can also write above equations as:
  & \dfrac{{{x}^{2}}}{4}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(5) \\
 & \dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\
Add both the equations (5) & (6), we get:
  & \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right) \\
 & +\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right) \\
 & ={{\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)}^{2}}.......(7)
Since \[\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)=1\], equation (7) becomes:
Equation (8) can also be written as: \[\dfrac{{{x}^{2}}}{8}+\dfrac{{{y}^{2}}}{18}=1......(9)\]

Compare equation (8) with the standard form of various conic sections.
We get to know that equation (8) represents ellipse whose standard form of equation is \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\]
Hence, \[\dfrac{{{x}^{2}}}{8}+\dfrac{{{y}^{2}}}{18}=1\] is an ellipse, where h=k=0; a= $ \sqrt{8} $ and b= $ \sqrt{18} $ .
Since a < b, the major axis of ellipse is parallel to the y-axis.
The ellipse represented by equation (8) is shown with the help of a diagram below. Here BD is the major axis and AC is the minor axis.
The foci of the ellipse lies on the major axis which is the y-axis itself. Those are represented by E $ \left( 0,be \right) $ and F $ \left( 0,-be \right) $ where ‘e’ is the eccentricity of the ellipse.
seo images

Now to find the foci of the ellipse, we need to calculate the eccentricity (e) of the ellipse.
i.e. for ellipse \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\];
If a < b, then $ e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}} $ $ e=\sqrt{1-\dfrac{{{a}^{2}}}{{{b}^{2}}}} $
If a > b, then $ e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}} $
Therefore, eccentricity of the given ellipse \[\dfrac{{{x}^{2}}}{8}+\dfrac{{{y}^{2}}}{18}=1\] is \[e=\sqrt{1-\dfrac{8}{18}}\]
\[e=\sqrt{\dfrac{2\times 5}{2\times 9}}\]
So focii of given ellipse are \[\left( 0,\sqrt{18}\times \dfrac{\sqrt{5}}{3} \right)\] and \[\left( 0,-\sqrt{18}\times \dfrac{\sqrt{5}}{3} \right)\]
i.e. \[(0,+\sqrt{10})\]and \[(0,-\sqrt{10})\]
So, the correct answer is “Option D”.

Note: Be careful while identifying the major axis of ellipse while considering the values of a and b because it might change the value of eccentricity and therefore foci would lie parallel to the x-axis in another case (i.e. a > b).
Suppose we subtract the square of both the equations instead of adding them to simplify and get an equation in terms of x and y.