Name: Tangent to a Hyperbola
Uploaded: 2019-07-19
Duration: 2 min 11 s
Description: The tangent to a plane curve at a given point is the straight line that "just touches" the curve at that point. Learn about the tangent to a hyperbola by wat...

Question

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola:

\frac{x^{2}}{100} - \frac{y^{2}}{25} = + 1

.

Accepted Answer

Given equation of the hyperbola: $\frac{x^{2}}{100} - \frac{y^{2}}{25} = 1$

The above equation represents the standard hyperbola(Transverse hyperbola) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

$∴ a^{2} = 100, b^{2} = 25 \Rightarrow a = 10, b = 5$ and $b^{2} = a^{2} (e^{2} - 1)$

$∴ 25 = 100 (e^{2} - 1)$

$\Rightarrow \frac{25}{100} = (e^{2} - 1)$

$\Rightarrow \frac{125}{100} = e^{2}$

$\Rightarrow e^{2} = \frac{5}{4}$

$\Rightarrow e = \frac{\sqrt{5}}{2} [∵ e > 0]$

Now,

(i) lengths of the principal axes:

$(1) \Rightarrow$ length of the transverse axis $2 a = 2 \times 10 = 20$

$(2) \Rightarrow$ length of the conjugate axis $2 b = 2 \times 5 = 10$

(ii) The eccentricity of the hyperbola $e = \frac{\sqrt{5}}{2}$

(iii) The foci of the hyperbola $(\pm a e, 0) = (\pm 10 \times \frac{\sqrt{5}}{2}, 0) = (\pm 5 \sqrt{5}, 0)$

(iv)The equation of directrices is $x = \pm \frac{a}{e} = \pm \frac{10}{\frac{\sqrt{5}}{2}} = \pm \frac{20}{\sqrt{5}}$

$\Rightarrow \sqrt{5} x \pm 20 = 0$

(v) The length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 25}{10} = 5$ .

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola: $\frac{x^{2}}{100} - \frac{y^{2}}{25} = + 1$ .

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola: x2100-y225=+1.

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola: $\frac{x^{2}}{100} - \frac{y^{2}}{25} = + 1$ .