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Question: How do u find the center, vertices, foci and asymptotes of \( \dfrac{{{x^2}}}{7} - \dfrac{{{y^2}}}{9...

How do u find the center, vertices, foci and asymptotes of x27y29=1\dfrac{{{x^2}}}{7} - \dfrac{{{y^2}}}{9} = 1 ?

Explanation

Solution

Hint : We should know about the equation of hyperbola before stat solving the question.
The standard form of equation of a hyperbola with center (c,d)(c,d) and transverse axis on the x-axis.
(xc)2a2(yd)2b2=1\dfrac{{{{(x - c)}^2}}}{{{a^2}}} - \dfrac{{{{(y - d)}^2}}}{{{b^2}}} = 1
Where,
The length of the transverse axis is 2a2a .
And, the length of the conjugate axis is the conjugate axis.

Complete step-by-step answer :
Step 1:
We make given equation simpler for us by writing in square of number to denominator values:
x27y232=1\dfrac{{{x^2}}}{{\sqrt 7 }} - \dfrac{{{y^2}}}{{{3^2}}} = 1
Step 2:
Compare it with standard equation:
(xc)2a2(yd)2b2=1\dfrac{{{{(x - c)}^2}}}{{{a^2}}} - \dfrac{{{{(y - d)}^2}}}{{{b^2}}} = 1
We get,
Centre C=(0,0)C = (0,0)
The vertices are V=(a,0)=(7,0)V' = ( - a,0) = ( - \sqrt 7 ,0) and V=(a,0)=(7,0)V' = (a,0) = (\sqrt 7 ,0) .
To calculate the foci, we need the distance from the centre to the foci,
c2=a2+b2{c^2} = {a^2} + {b^2}
c2=7+9=16\Rightarrow {c^2} = 7 + 9 = 16
c=±4\Rightarrow c = \pm 4
The foci are F=(c,0)=(4,0)F' = ( - c,0) = ( - 4,0) and F=(c,0)=(4,0)F = (c,0) = (4,0) .
The asymptotes will be,
x27y232=1\dfrac{{{x^2}}}{{\sqrt 7 }} - \dfrac{{{y^2}}}{{{3^2}}} = 1
y=±37y = \pm \dfrac{3}{{\sqrt 7 }} .

Note : As we come through many applications of hyperbola in our daily life. A guitar is an example of a hyperbola as its sides form a hyperbola. Airport design is hyperbolic parabolic. As it has one cross section of hyperbola and others have a parabola. Gear transmission having pair of hyperbolic gear. Hyperbola is important in astronomy as they are the path followed by the non-recurrent comets.