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

At which point on the curve $3x^2 - y^2 = 8$ is the normal parallel to the line $x + 3y = 4$?

The equation of the tangent at the point $(a \sec \theta, b \tan \theta)$ of the conic $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

The equation of the normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ that is perpendicular to the line $2x + y = 1$ is:

If $e$ and $e'$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,then $\frac{1}{e^2} + \frac{1}{e'^2} = \dots$

If the ratio of the distance between the foci to the distance between the directrices of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $3 : 2$,then $a : b = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module