Question

Which one of the following equations represents the parametric equation to a parabolic curve?
A) $x = 3\cos t;y = 4\sin t$
B) ${x^2} - 2 = 2\cos t;y = 4{\cos ^2}\dfrac{t}{2}$
C) $\sqrt x = \tan t;\sqrt y = \sec t$
D) $x = \sqrt {1 - \sin t} ;y = \sin \dfrac{t}{2} + \cos \dfrac{t}{2}$

Answer 1

Here, it is asked to find the parametric equation of a parabola with the help of information provided in options.
So, make equations using the information from the options and write down the type of shape represented by the equation.
Thus, choose the required answer as the correct option.

Complete step by step solution:
Here, we are asked to find the parametric equation of a parabola.
To do that, we will check all of the given options on-by-one and see which option can be transformed in the format of ${x^2} = 4ay$, which is the general equation of a parabola.
So, in option (A) the given data is $x = 3\cos t;y = 4\sin t$.
Now, on squaring $y = 4\sin t$ gives ${\left( y \right)^2} = {\left( {4\sin t} \right)^2}$ .
$\Rightarrow {y^2} = 16{\sin ^2}t$
Also, ${\sin ^2}t = 1 - {\cos ^2}t$
$\Rightarrow {y^2} = 16\left( {1 - {{\cos }^2}t} \right)$
$\Rightarrow \dfrac{{{y^2}}}{{16}} = 1 - {\cos ^2}t$ ... (1)
Now, squaring $x = 3\cos t$ gives ${\left( x \right)^2} = {\left( {3\cos t} \right)^2}$
$\Rightarrow {x^2} = 9{\cos ^2}t \Rightarrow {\cos ^2}t = \dfrac{{{x^2}}}{9}$
So, substituting the value of ${\cos ^2}t$ in equation (1) gives
$\dfrac{{{y^2}}}{{16}} = 1 - \dfrac{{{x^2}}}{9}$
$\Rightarrow \dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = 1$
Thus, we get the equation of an ellipse. So, option (A) is not the required answer.
Now, in option (B) the given data is ${x^2} - 2 = 2\cos t;y = 4{\cos ^2}\dfrac{t}{2}$ .
We can write ${x^2} - 2 = 2\cos t$ as ${x^2} = 2 + 2\cos t$.
$\Rightarrow {x^2} = 2\left( {1 + \cos t} \right)$
Also, $1 + \cos t = 2{\cos ^2}\dfrac{t}{2}$
$\Rightarrow {x^2} = 2 \times 2{\cos ^2}\dfrac{t}{2}$
$\Rightarrow {x^2} = 4{\cos ^2}\dfrac{t}{2}$
$\Rightarrow {x^2} = y\left( {\because y = 4{{\cos }^2}t} \right)$
Thus, we get the equation of a parabola. So, option (B) is the required answer.
Let us check further and verify whether the given question has multiple correct answers or not.
In option (C) the given data is $\sqrt x = \tan t;\sqrt y = \sec t$.
Now, squaring both sides in $\sqrt y = \sec t$
$\Rightarrow {\left( {\sqrt y } \right)^2} = {\left( {\sec t} \right)^2} \\\ \Rightarrow y = {\sec ^2}t$
Also, ${\sec ^2}t = 1 + {\tan ^2}t$
$\Rightarrow y = 1 + {\tan ^2}t$
It is also given that $\sqrt x = \tan t \Rightarrow x = {\tan ^2}t$ (squaring both sides).
$\Rightarrow y = 1 + x$
Thus, we get the equation of a straight line. So, option (C) is also not the required answer.
In option (D), the given data is $x = \sqrt {1 - \sin t} ;y = \sin \dfrac{t}{2} + \cos \dfrac{t}{2}$ .
So, squaring both sides is $x = \sqrt {1 - \sin t} ;y = \sin \dfrac{t}{2} + \cos \dfrac{t}{2}$ , gives
${\left( x \right)^2} = {\left( {\sqrt {1 - \sin t} } \right)^2}$
$\Rightarrow {x^2} = 1 - \sin t \\\ \Rightarrow \sin t = 1 - {x^2}$
And

$${\left( y \right)^2} = {\left( {\sin \dfrac{t}{2} + \cos \dfrac{t}{2}} \right)^2} \\\ \Rightarrow {y^2} = {\sin ^2}\dfrac{t}{2} + {\cos ^2}\dfrac{t}{2} + 2\sin \dfrac{t}{2}\cos \dfrac{t}{2}$$

Now, ${\sin ^2}\dfrac{t}{2} + {\cos ^2}\dfrac{t}{2} = 1$ and $2\sin \dfrac{t}{2}\cos \dfrac{t}{2} = \sin 2\left( {\dfrac{t}{2}} \right) = \sin t$
$\Rightarrow {y^2} = 1 + \sin t \\\ \Rightarrow {y^2} = 1 + 1 - {x^2}\left( {\because \sin t = 1 - {x^2}} \right) \\\ \Rightarrow {x^2} + {y^2} = 2$
Thus, we get the equation of the circle. So, option (D) is also not the required answer.

Hence, we get option (C) as our required answer.

Note:
The general equations for some of the below given geometric shapes is as follows:
Eclipse: The equation of ellipse a and b as the radius on the X and Y axes respectively is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.
Circle: The equation of circle with radius r is ${x^2} + {y^2} = {r^2}$.
Parabola: The equation for the parabola symmetric about X-axis is ${x^2} = 4ay$ and that for the parabola symmetric about Y-axis is ${y^2} = 4bx$.
Hyperbola: The standard form of the equation of a hyperbola with centre $\left( {0,0} \right)$ and transverse axis on the X-axis is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. The standard form of the equation of a hyperbola with centre $\left( {0,0} \right)$ and transverse axis on the Y-axis is $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$.
Straight line: The equation of straight line with slope m and intercept c is $y = mx + c$.