Question

# 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 el

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

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 $$\displaystyle\frac{{x}^{2}}{{52}}+\frac{{y}^{2}}{{13}}={1}$$.