Question

Find the equation of the hyperbola where foci are ( 0+-12 ) and the length of the latus rectum is 36

Answer 1

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Answer 2

Answer:

The equation of the hyperbola is

$\frac{y^2}{36}-\frac{x^2}{108}$=1

Step-by-step explanation:

As per the data given in the question we have to find the equation of the hyperbola.

As per the questions it is given that foci are ( 0+-12 ) and the length of the latus rectum is 36.

The word ‘latus rectum’ is derived from the Latin word ‘latus’ which means ‘side’ and the rectum which means ‘straight’. The latus rectum is defined as the chord passing through the focus, and perpendicular to the directrix. The end point of the latus rectum lies on the curve. Half of the latus rectum is considered as the semi latus rectum. The length of the latus rectum of each conic section is defined differently.

The Latus rectum of a hyperbola is defined as a line segment perpendicular to the transverse axis through any of the foci and whose ending point lies on the hyperbola. The length of the latus rectum of a hyperbola is 2b²/a.

On a hyperbola, focus (foci being plural) are the fixed points such that the difference between the distances are always found to be constant.

Thus the value of ae=12 and $\frac{2b^2}{a}$ = 36

⇒$b^{2} =$18a

Now by using the relation

$b^{2}=a^{2}(e^{2}-1)$we get as

⇒$18a=144-a^{2}$

⇒$a^{2}$+18a-144=0

⇒(a+24)(a-6) =0

⇒a=-24 and 6

$b^{2}$ = -432 or 108 (negative value is not possible)

Thus the equation of hyperbola is

$\frac{y^2}{36}-\frac{x^2}{108}$=1

Hence, the equation of the hyperbola is

$\frac{y^2}{36}-\frac{x^2}{108}$=1

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