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

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

Ezra Herbert

Skilled2020-11-11Added 99 answers

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,

$\frac{{4}^{2}}{{a}^{2}}+\frac{{3}^{2}}{{b}^{2}}=1$

$\Rightarrow \frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}=1$

$\Rightarrow \frac{16}{{a}^{2}}=1-\frac{9}{{b}^{2}}$

$\Rightarrow \frac{1}{{a}^{2}}=\frac{1}{16}(1-\frac{9}{{b}^{2}})$

Step 2

Equation of the ellipse through the point (6, 2).

So,$x=6,y=2$ satisfies the equation of an ellipse.

$\Rightarrow \frac{{6}^{2}}{{a}^{2}}+\frac{{2}^{2}}{{b}^{2}}=1$

$\Rightarrow \frac{36}{{a}^{2}}+\frac{4}{{b}^{2}}=1$

$\Rightarrow \frac{36}{{a}^{2}}=1-\frac{4}{{b}^{2}}$

$\Rightarrow \frac{1}{{a}^{2}}=\frac{1}{36}(1-\frac{4}{{b}^{2}})$

Step 3

Compare equations (1) and (2), we get

$\frac{1}{36}(1-\frac{4}{{b}^{2}})=\frac{1}{16}(1-\frac{9}{{b}^{2}})$

Multiply both sides by 4,

$\frac{1}{9}(1-\frac{4}{{b}^{2}})=\frac{1}{4}(1-\frac{9}{{b}^{2}})$

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,

$\frac{4}{36}-\frac{9}{36}=\frac{16}{36{b}^{2}}-\frac{81}{36{b}^{2}}$

$\Rightarrow -\frac{5}{36}=-\frac{65}{36{b}^{2}}$

$\Rightarrow -\frac{5}{36}=-\frac{65}{36{b}^{2}}$

Multiply both sides by 36,

$\Rightarrow -\frac{5}{1}=-\frac{65}{{b}^{2}}$

Divide both sides by -65,

$\Rightarrow \frac{5}{65}=\frac{1}{{b}^{2}}$

By simplifying it,

$\Rightarrow \frac{1}{13}=\frac{1}{{b}^{2}}$

Taking reciprocal from both sides,

$\Rightarrow {b}^{2}=13$

Step 4

Substitute${b}^{2}=13$

in the equation$\frac{1}{{a}^{2}}=\frac{1}{16}(1-\frac{9}{{b}^{2}}),$

An equation of the ellipse is of the form,

Equation of the ellipse through the point (4, 3).

So,

Implies that,

Step 2

Equation of the ellipse through the point (6, 2).

So,

Step 3

Compare equations (1) and (2), we get

Multiply both sides by 4,

Distribute both sides,

Combine like terms

Make the same denominator,

Multiply both sides by 36,

Divide both sides by -65,

By simplifying it,

Taking reciprocal from both sides,

Step 4

Substitute

in the equation