Question

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)$(0,\pm \sqrt{10})$

Answer 1

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{aligned} & y=\dfrac{{{\left( x-h \right)}^{2}}}{4a}+k \\\ & x=\dfrac{{{\left( y-k \right)}^{2}}}{4a}+h \\\ \end{aligned} \right\\}
Ellipse: $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$
Hyperbola: \left\\{ \begin{aligned} & \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{aligned} \right\\}$$$$\left\\{ \begin{aligned} & \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{aligned} \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) \\\ \end{aligned}$$ Squaring the both sides of equation (1) & (2), By applying the identities:$$\left[ \begin{aligned} & \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \because {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ \end{aligned} \right]$$$$\left[ \begin{aligned} & \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \because {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ \end{aligned} \right]$$ we get: $$\begin{aligned} & {{x}^{2}}=4\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(3) \\\ & {{y}^{2}}=9\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\\ \end{aligned}$$ We can also write above equations as: $$\begin{aligned} & \dfrac{{{x}^{2}}}{4}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right)......(5) \\\ & \dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right)......(4) \\\ \end{aligned}$$ Add both the equations (5) & (6), we get: $$\begin{aligned} & \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t}-2\sin \text{t}\text{.}\cos \text{t} \right) \\\ & +\left( {{\cos }^{2}}\text{t}+{{\sin }^{2}}\text{t }+\text{ }2\sin \text{t}\text{.}\cos \text{t} \right) \\\ & ={{\left( {{\cos }^{2}}\text{t +}{{\sin }^{2}}\text{t} \right)}^{2}}.......(7) \end{aligned}$$ 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.