Question

Eliminate theta
X=3sec ^,y=4tan^

Answer 1

Answer:

$\bf\frac{x^2}{9}-\frac{y^2}{16}=1$

Step-by-step explanation:

Given:

$x=3\:sec\theta$ and

$y=4\:tan\theta$

$\implies\:\frac{x}{3}=sec\theta\:\text{and}\:\frac{y}{4}=tan\theta$

we know that

$\boxed{sec^2\theta-tan^2\theta=1}$

$\implies\:(\frac{x}{3})^2-(\frac{y}{4})^2=1$

$\implies\:\bf\frac{x^2}{9}-\frac{y^2}{16}=1$

Answer 2

Answer :

$16 {x}^{2} - 9 {y}^{2} = 1$

Step by step Explanation :

$x = 3 \sec\theta \: and \: 9 \tan\theta$

$\therefore \frac{x}{3} = \sec \theta \: and \: \frac{y}{9} = \tan \theta$

On squaring and subtracting on both sides,

$\frac{ {x}^{2} }{3 ^{2} } - \frac{ {y}^{2} }{ {4}^{2} } = \sec^{2}\theta - \tan^{2} \theta$

$\frac{ {x}^{2} }{9} - \frac{ {y}^{2} }{16 } = \sec^{2} \theta - \tan^{2} \theta$

$but \: \sec^{2} - { \tan}^{2} = 1$

$\frac{ {x}^{2} }{9} - \frac{ {y}^{2} }{16} = 1$

Now the final Answer will be,

$16 {x}^{2} - 9 {y}^{2} = 1$