 kuCAu

2020-11-10

Solve for the equation in standard form of the following conic sections and graph the curve on a Cartesian plane indicating important points.
1. An ellipse passing through (4, 3) and (6, 2)
2. A parabola with axis parallel to the x-axis and passing through (5, 4), (11, -2) and (21, -4)
3. The hyperbola given by $5{x}^{2}-4{y}^{2}=20x+24y+36.$ Ezra Herbert

Step 1
An equation of the ellipse is of the form,
$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
Equation of the ellipse through the point (4, 3).
So, $x=4,y=3$ satisfies the equation of an ellipse.
Implies that,
$\frac{{4}^{2}}{{a}^{2}}+\frac{{3}^{2}}{{b}^{2}}=1$
$⇒\frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}=1$
$⇒\frac{16}{{a}^{2}}=1-\frac{9}{{b}^{2}}$
$⇒\frac{1}{{a}^{2}}=\frac{1}{16}\left(1-\frac{9}{{b}^{2}}\right)$
Step 2
Equation of the ellipse through the point (6, 2).
So, $x=6,y=2$ satisfies the equation of an ellipse.
$⇒\frac{{6}^{2}}{{a}^{2}}+\frac{{2}^{2}}{{b}^{2}}=1$
$⇒\frac{36}{{a}^{2}}+\frac{4}{{b}^{2}}=1$
$⇒\frac{36}{{a}^{2}}=1-\frac{4}{{b}^{2}}$
$⇒\frac{1}{{a}^{2}}=\frac{1}{36}\left(1-\frac{4}{{b}^{2}}\right)$
Step 3
Compare equations (1) and (2), we get
$\frac{1}{36}\left(1-\frac{4}{{b}^{2}}\right)=\frac{1}{16}\left(1-\frac{9}{{b}^{2}}\right)$
Multiply both sides by 4,
$\frac{1}{9}\left(1-\frac{4}{{b}^{2}}\right)=\frac{1}{4}\left(1-\frac{9}{{b}^{2}}\right)$
Distribute both sides,
$\frac{1}{9}-\frac{4}{9{b}^{2}}=\frac{1}{4}-\frac{9}{4{b}^{2}}$
Combine like terms
$\frac{1}{9}-\frac{1}{4}=\frac{4}{9{b}^{2}}-\frac{9}{4{b}^{2}}$
Make the same denominator,
$\frac{4}{36}-\frac{9}{36}=\frac{16}{36{b}^{2}}-\frac{81}{36{b}^{2}}$
$⇒-\frac{5}{36}=-\frac{65}{36{b}^{2}}$
$⇒-\frac{5}{36}=-\frac{65}{36{b}^{2}}$
Multiply both sides by 36,
$⇒-\frac{5}{1}=-\frac{65}{{b}^{2}}$
Divide both sides by -65,
$⇒\frac{5}{65}=\frac{1}{{b}^{2}}$
By simplifying it,
$⇒\frac{1}{13}=\frac{1}{{b}^{2}}$
Taking reciprocal from both sides,
$⇒{b}^{2}=13$
Step 4
Substitute ${b}^{2}=13$
in the equation $\frac{1}{{a}^{2}}=\frac{1}{16}\left(1-\frac{9}{{b}^{2}}\right),$

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