Question
The curve described parametrically by $$x = 2 - 3\,\sec \,t,\,y = 1 + 4\,\tan \,t$$ represents :
A.
An ellipse centered at $$\left( {2,\,1} \right)$$ and of eccentricity $$\frac{3}{5}$$
B.
A circle centered at $$\left( {2,\,1} \right)$$ and of radius $$5$$ units
C.
A hyperbola centered at $$\left( {2,\,1} \right)$$ & of eccentricity $$\frac{8}{5}$$
D.
A hyperbola centered at $$\left( {2,\,1} \right)$$ & of eccentricity $$\frac{5}{3}$$
Answer :
A hyperbola centered at $$\left( {2,\,1} \right)$$ & of eccentricity $$\frac{5}{3}$$
Solution :
Given, $$x = 2 - 3\,\sec \,t,\,y = 1 + 4\,\tan \,t$$
$$ \Rightarrow \sec \,t = \frac{{x - 2}}{{ - 3}},\,\tan \,t = \frac{{y - 1}}{4}$$
Since, $${\sec ^2}t - {\tan ^2}t = 1$$
$$\therefore \,\frac{{{{\left( {x - 2} \right)}^2}}}{9} - \frac{{{{\left( {y - 1} \right)}^2}}}{{16}} = 1$$
which is a hyperbola with centre at $$\left( {2,\,1} \right)$$ and eccentricity $$e,$$ given by
$$\eqalign{
& 16 = 9\left( {{e^2} - 1} \right) \cr
& \Rightarrow 9{e^2} = 25 \cr
& \Rightarrow {e^2} = \frac{{25}}{9} \cr
& \Rightarrow e = \frac{5}{3} \cr} $$