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

(i) lengths of the principal axes

(ii) co-ordinates of the foci

(iii) equations of directrices

( iv) length of the latus rectum

(v) Distance between foci

(vi) distance between directrices of the curve

$\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$

Accepted Answer

Given equation of the hyperbola is $\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$ .

Comparing with the standard equation: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ we get,

$a^{2} = 144 and b^{2} = 25 \Rightarrow a = 12 and b = 5$

Here, $(a > b)$ , so $x$ -axis $(y = 0)$ is the major axis and $y$ -axis $(x = 0)$ is the minor axis.

(i) We know that, length of the major axis is $2 a$ and length of the minor axis is $2 b$ .

Therefore, length of the major axis is $2 \times 12 = 24$ .

Length of the minor axis is $2 \times 5 = 10$

(ii) Let the eccentricity of the hyperbola is $e$ .

We know that, $b^{2} = a^{2} (e^{2} - 1)$

$\Rightarrow 25 = 144 (e^{2} - 1)$

$\Rightarrow \frac{25}{144} = e^{2} - 1$

$\Rightarrow e^{2} = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$

$\Rightarrow e = \pm \frac{13}{12}$

Now we know that the coordinates of the foci are $(a e, 0)$ .

Therefore, the foci is $(12 \times (\pm \frac{13}{12}), 0) = (\pm 13, 0)$ .

(iii) Equation of directrix is $x = \frac{a}{e} = \pm \frac{12}{\frac{13}{12}} = \pm \frac{144}{13}$

Therefore, the equation of the directrix is $x = \pm \frac{144}{12}$ .

(iv) Distance between foci $= 2 a e = 2 \times 12 \times \frac{13}{12} = 26$

(v) Distance between directrix $= 2 \times \frac{a}{e} = \frac{2 \times 12 \times 12}{13} = \frac{288}{13}$

(vi) Length of the latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 25}{12} = \frac{25}{6}$

Find the

(i) lengths of the principal axes

(ii) co-ordinates of the foci

(iii) equations of directrices

( iv) length of the latus rectum

(v) Distance between foci

(vi) distance between directrices of the curve

$\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Find the (i) lengths of the principal axes (ii) co-ordinates of the foci (iii) equations of directrices ( iv) length of the latus rectum (v) Distance between foci (vi) distance between directrices of the curve x2144-y225=1

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Find the

(i) lengths of the principal axes

(ii) co-ordinates of the foci

(iii) equations of directrices

( iv) length of the latus rectum

(v) Distance between foci

(vi) distance between directrices of the curve

$\frac{x^{2}}{144} - \frac{y^{2}}{25} = 1$