Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\,\pm 3),$ foci $(0,\,±5)$
Vertices $(0,\,\pm 3),$ foci $(0,\,±5)$
Here, the vertices are on the $y-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
since the vertices are $(0,\,\pm 3), a=3$
since the foci are $(0,\,\pm 5), c=5$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore 3^{2}+b^{2}=52$
$\Rightarrow b^{2}=25-9=16$
Thus, the equation of the hyperbola is $\frac{y^{2}}{9}-\frac{x^{2}}{16}=1$
Let $P \left( x _0, y _0\right)$ be the point on the hyperbola $3 x ^2-4 y ^2$ $=36$, which is nearest to the line $3 x+2 y=1$. Then $\sqrt{2}\left( y _0- x _0\right)$ is equal to :
A hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3 x^{2}+4 y^{2}=12,$ then this hyperbola does not pass through which of the following points?
If $(0,\; \pm 4)$ and $(0,\; \pm 2)$ be the foci and vertices of a hyperbola, then its equation is
Equations of a common tangent to the two hyperbolas $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}$ $= 1 $ $\&$ $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}}$ $= 1 $ is :
The locus of the point of intersection of the lines $ax\sec \theta + by\tan \theta = a$ and $ax\tan \theta + by\sec \theta = b$, where $\theta $ is the parameter, is