The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is
The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is
- A
$5$
- B
$7$
- C
$9$
- D
$4$
Similar Questions
Find the equation of the hyperbola satisfying the give conditions: Foci $(0, \,\pm \sqrt{10}),$ passing through $(2,\,3)$
Find the equation of the hyperbola satisfying the give conditions: Foci $(0, \,\pm \sqrt{10}),$ passing through $(2,\,3)$
The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is
The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is
What is the slope of the tangent line drawn to the hyperbola $xy = a\,(a \ne 0)$ at the point $(a, 1)$
What is the slope of the tangent line drawn to the hyperbola $xy = a\,(a \ne 0)$ at the point $(a, 1)$
If line $ax$ + $by$ = $1$ is normal to the hyperbola $\frac{{{x^2}}}{{{p^2}}} - \frac{{{y^2}}}{{{q^2}}} = 1$ then $\frac{{{p^2}}}{{{a^2}}} - \frac{{{q^2}}}{{{b^2}}} = 1$ is equal to (where $a$,$b$,$p$, $q \in {R^ + })$-
If line $ax$ + $by$ = $1$ is normal to the hyperbola $\frac{{{x^2}}}{{{p^2}}} - \frac{{{y^2}}}{{{q^2}}} = 1$ then $\frac{{{p^2}}}{{{a^2}}} - \frac{{{q^2}}}{{{b^2}}} = 1$ is equal to (where $a$,$b$,$p$, $q \in {R^ + })$-
The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is
The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is