MCQ
If $x=a \tan \theta$ and $y=b$ eec $\theta$, then
- ✓$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
- B$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
- C$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
- D$\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
✓
Answer
Correct option: A.
$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
(a) : We have, $x=a \tan \theta$ and $y=b \sec \theta$
$\Rightarrow \tan \theta=\frac{x}{2}$ and $\sec \theta=\frac{z}{b}$
Putting these values in $\sec ^2 \theta-\tan ^2 \theta=1$, we got
$
\frac{y^2}{ b ^2}-\frac{ x ^2}{a^2}=1
$
$\Rightarrow \tan \theta=\frac{x}{2}$ and $\sec \theta=\frac{z}{b}$
Putting these values in $\sec ^2 \theta-\tan ^2 \theta=1$, we got
$
\frac{y^2}{ b ^2}-\frac{ x ^2}{a^2}=1
$
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