Question

$x=4\left(1+\cos \theta \right)$ and $y=3\left(1+\sin \theta \right)$ are the parametric equations of

• A. $\left[\frac{{\left(x-3\right)}^2}{9}\right]+\left[\frac{{\left(y-4\right)}^2}{16}\right]=1$
• B. $\left[\frac{{\left(x+4\right)}^2}{16}\right]+\left[\frac{{\left(y+3\right)}^2}{9}\right]=1$
• C. $\left[\frac{{\left(x-4\right)}^2}{16}\right]-\left[\frac{{\left(y-3\right)}^2}{9}\right]=1$
• D. $\left[\frac{{\left(x-4\right)}^2}{16}\right]+\left[\frac{{\left(y-3\right)}^2}{9}\right]=1$

Answer 1

The correct option is D

$\left[\frac{{\left(x-4\right)}^2}{16}\right]+\left[\frac{{\left(y-3\right)}^2}{9}\right]=1$

Explanation for the correct option:

Step 1: Note the given data

$x=4\left(1+\cos \theta \right)..........\left(1\right)$

$y=3\left(1+\sin \theta \right)..........\left(2\right)$

A parametric function is a function with a one-dimensional input and a multidimensional output.

Step 2: Find the value of $\cos \theta$ and $\sin \theta$

$x=4+4\cos \theta$

$\Rightarrow \cos \theta =\frac{x-4}{4}$

And $y=3+3\sin \theta$

$\Rightarrow \sin \theta =\frac{y-3}{3}$

We know the trigonometric relation ${\cos }^2\theta +{\sin }^2\theta =1$

Substituting the values of $\cos \theta$ and $\sin \theta$

$\Rightarrow {\left(\frac{x-4}{4}\right)}^2+{\left(\frac{y-3}{3}\right)}^2=1$

$\Rightarrow \left[{\frac{\left(x-4\right)}{16}}^2\right]+\left[{\frac{\left(y-3\right)}{9}}^2\right]=1$

Hence, the correct option is D.