If the centre,vertex,and focus of a hyperbola | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: Centre $(h, k) = (0, 0)$,Vertex $(a, 0) = (4, 0)$,and Focus $(ae, 0) = (6, 0)$.From the vertex,$a = 4$.From the focus,$ae = 6$. Substitutin

Vedclass · Accepted Answer

$5x^2 - 4y^2 = 80$

Vedclass · Answer

(C) Given: Centre $(h, k) = (0, 0)$,Vertex $(a, 0) = (4, 0)$,and Focus $(ae, 0) = (6, 0)$.From the vertex,$a = 4$.From the focus,$ae = 6$. Substituting $a = 4$,we get $4e = 6$,so $e = \frac{6}{4} = \frac{3}{2}$.For a hyperbola,$b^2 = a^2(e^2 - 1)$.$b^2 = 16 \left( (\frac{3}{2})^2 - 1 \right) = 16 (\frac{9}{4} - 1) = 16 (\frac{5}{4}) = 20$.The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Substituting $a^2 = 16$ and $b^2 = 20$,we get $\frac{x^2}{16} - \frac{y^2}{20} = 1$.Multiplying by $80$,we get $5x^2 - 4y^2 = 80$.

If the centre,vertex,and focus of a hyperbola are $(0, 0)$,$(4, 0)$,and $(6, 0)$ respectively,then the equation of the hyperbola is

Similar Questions

If the equation of one asymptote of the hyperbola $14 x^2+38 x y+20 y^2+x-7 y-91=0$ is $7 x+5 y-3=0$,then the other asymptote is

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

If $\frac{x^2}{k-\frac{5}{2}}+\frac{y^2}{\frac{7}{3}-k}= 1$ ($k$ is a real number) represents a hyperbola,then the set of all values of $k$ is

The equation $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module