Let the X-axis be the transverse axis and the | vedclass

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

Question

(A) The equation of the director circle of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2 - b^2$.Given the director circle is

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The equation of the director circle of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2 - b^2$.Given the director circle is $x^2 + y^2 = 16$,we have $a^2 - b^2 = 16$  $(i)$.The perpendicular distance from the centre $(0,0)$ to the latus rectum of the hyperbola is given by $\frac{a^2}{c}$ or simply the distance to the line $x = ae$,which is $ae$.Given $ae = \sqrt{34}$,then $c^2 = a^2 + b^2 = 34$  $(ii)$.Adding $(i)$ and $(ii)$: $(a^2 - b^2) + (a^2 + b^2) = 16 + 34$ $\Rightarrow 2a^2 = 50$ $\Rightarrow a^2 = 25$ $\Rightarrow a = 5$.Substituting $a^2 = 25$ into $(ii)$: $25 + b^2 = 34$ $\Rightarrow b^2 = 9$ $\Rightarrow b = 3$.Therefore,$a + b = 5 + 3 = 8$.

Let the $X$-axis be the transverse axis and the $Y$-axis be the conjugate axis of a hyperbola $H$. Let $x^2+y^2=16$ be the director circle of $H$. If the perpendicular distance from the centre of $H$ to its latus rectum is $\sqrt{34}$,then $a+b=$

Similar Questions

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $5y^{2} - 9x^{2} = 36$.

The equation of the tangent to the hyperbola $4y^2 = x^2 - 1$ at the point $(1, 0)$ is

The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} - 3{y^2} = 1$ is

If $(0, \pm 4)$ and $(0, \pm 2)$ are the foci and vertices of a hyperbola,respectively,then its equation is:

If the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{5}{4}$ and $2x+3y-6=0$ is a focal chord of the hyperbola,then the length of the transverse axis is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module