Answer
425.1k+ views
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:
\[\begin{align}
& x=2\left( \cos \text{t}-\sin \text{t} \right)......(1) \\
& y=3\left( \cos \text{t}+\sin \text{t} \right)......(2) \\
\end{align}\]
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:
\[\begin{align}
& {{x}^{2}}=4\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(3) \\
& {{y}^{2}}=9\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\
\end{align}\]
We can also write above equations as:
\[\begin{align}
& \dfrac{{{x}^{2}}}{4}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(5) \\
& \dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\
\end{align}\]
Add both the equations (5) & (6), we get:
\[\begin{align}
& \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right) \\
& +\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right) \\
& ={{\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)}^{2}}.......(7)
\end{align}\]
Since \[\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)=1\], equation (7) becomes:
\[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=2......(8)\]
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.
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 \[e=\sqrt{1-\dfrac{8}{18}}\]
\[e=\sqrt{\dfrac{10}{18}}\]
\[e=\sqrt{\dfrac{2\times 5}{2\times 9}}\]
\[e=\dfrac{\sqrt{5}}{3}\]
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.
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:
\[\begin{align}
& x=2\left( \cos \text{t}-\sin \text{t} \right)......(1) \\
& y=3\left( \cos \text{t}+\sin \text{t} \right)......(2) \\
\end{align}\]
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:
\[\begin{align}
& {{x}^{2}}=4\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(3) \\
& {{y}^{2}}=9\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\
\end{align}\]
We can also write above equations as:
\[\begin{align}
& \dfrac{{{x}^{2}}}{4}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(5) \\
& \dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\
\end{align}\]
Add both the equations (5) & (6), we get:
\[\begin{align}
& \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right) \\
& +\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right) \\
& ={{\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)}^{2}}.......(7)
\end{align}\]
Since \[\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)=1\], equation (7) becomes:
\[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=2......(8)\]
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](https://www.vedantu.com/question-sets/1ff0439d-aaca-47d6-b2ab-846081104b0e4263166620381567499.png)
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 \[e=\sqrt{1-\dfrac{8}{18}}\]
\[e=\sqrt{\dfrac{10}{18}}\]
\[e=\sqrt{\dfrac{2\times 5}{2\times 9}}\]
\[e=\dfrac{\sqrt{5}}{3}\]
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)