Question

What is the equation of a curve given by the parametric form $x=9+6sec\theta ;y=-2-4tan\theta$.

• A. $\frac{(x+9{)}^2}{36}-\frac{(y-2{)}^2}{16}=1$
• B. $\frac{(x-9{)}^2}{36}-\frac{(y+2{)}^2}{16}=1$
• C. $\frac{(x-6{)}^2}{81}-\frac{(y+4{)}^2}{4}=1$
• D. $\frac{(x+6{)}^2}{81}-\frac{(y-4{)}^2}{4}=1$

Answer 1

The correct option is B

$\frac{(x-9{)}^2}{36}-\frac{(y+2{)}^2}{16}=1$

Given parametric equation are,

$x=9+6sec\theta$

$y=-2-4tan\theta$

Eliminating $\theta$ from these

$se{c}^2\theta -ta{n}^2\theta =1$

i.e., ${\left(\frac{x-9}{6}\right)}^2-{\left(\frac{y+2}{4}\right)}^2=1$

$\frac{(x-9{)}^2}{36}-\frac{(y+2{)}^2}{16}=1$