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 displaystyle{5}{x}^{2}-{4}{y}^{2}={20}{x}+{24}{y}+{36}.

Question
Conic sections
asked 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 \(\displaystyle{5}{x}^{2}-{4}{y}^{2}={20}{x}+{24}{y}+{36}.\)

Answers (1)

2020-11-11
Step 1
An equation of the ellipse is of the form,
\(\displaystyle\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}={1}\)
Equation of the ellipse through the point (4, 3).
So, \(\displaystyle{x}={4},{y}={3}\) satisfies the equation of an ellipse.
Implies that,
\(\displaystyle\frac{{4}^{2}}{{a}^{2}}+\frac{{3}^{2}}{{b}^{2}}={1}\)
\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}={1}\)
\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}={1}-\frac{9}{{b}^{2}}\)
\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)
Step 2
Equation of the ellipse through the point (6, 2).
So, \(\displaystyle{x}={6},{y}={2}\) satisfies the equation of an ellipse.
\(\displaystyle\Rightarrow\frac{{6}^{2}}{{a}^{2}}+\frac{{2}^{2}}{{b}^{2}}={1}\)
\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}+\frac{4}{{b}^{2}}={1}\)
\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}={1}-\frac{4}{{b}^{2}}\)
\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}\)
Step 3
Compare equations (1) and (2), we get
\(\displaystyle\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)
Multiply both sides by 4,
\(\displaystyle\frac{1}{{9}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{4}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)
Distribute both sides,
\(\displaystyle\frac{1}{{9}}-\frac{4}{{{9}{b}^{2}}}=\frac{1}{{4}}-\frac{9}{{{4}{b}^{2}}}\)
Combine like terms
\(\displaystyle\frac{1}{{9}}-\frac{1}{{4}}=\frac{4}{{{9}{b}^{2}}}-\frac{9}{{{4}{b}^{2}}}\)
Make the same denominator,
\(\displaystyle\frac{4}{{36}}-\frac{9}{{36}}=\frac{16}{{{36}{b}^{2}}}-\frac{81}{{{36}{b}^{2}}}\)
\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)
\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)
Multiply both sides by 36,
\(\displaystyle\Rightarrow-\frac{5}{{1}}=-\frac{65}{{b}^{2}}\)
Divide both sides by -65,
\(\displaystyle\Rightarrow\frac{5}{{65}}=\frac{1}{{b}^{2}}\)
By simplifying it,
\(\displaystyle\Rightarrow\frac{1}{{13}}=\frac{1}{{b}^{2}}\)
Taking reciprocal from both sides,
\(\displaystyle\Rightarrow{b}^{2}={13}\)
Step 4
Substitute \(\displaystyle{b}^{2}={13}\)
in the equation \(\displaystyle\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)},\)
\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{13}}\right)}\)
\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left(\frac{4}{{13}}\right)}\)
\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{4}}{\left(\frac{1}{{13}}\right)}\)
Taking reciprocal from both sides
\(\displaystyle\Rightarrow{a}^{2}={52}\)
The equation of ellipse through the points (4, 3) and (6, 2) is \(\displaystyle\frac{{x}^{2}}{{52}}+\frac{{y}^{2}}{{13}}={1}\).
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