## Solution of Exercise 6: Three conic lines (C7 – Math 10 Kite) – Math book

Solution of Exercise 6: Three conic lines (C7 – Math 10 Kite)
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### Solve the exercises Lesson 1 page 102 Math textbook 10 Kite episode 2

Which of the following is the canonical equation of the ellipse?

$$a)\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{64}} = 1$$

b) $$\frac{{{x^2}}}{{64}} – \frac{{{y^2}}}{{64}} = 1$$

c) $$\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{25}} = 1$$

d) $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{64}} = 1$$

Solution method

The ellipse (E) has the canonical equation: $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^) 2}}} = 1\left( {a > b > 0} \right)$$

Solution guide

The canonical equation of the ellipse is: c) $$\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{25}} = 1$$.

a) Not a PTCT because a = b = 8

b) Not a PTCT

d) Not a PTCT because a = 5 < b = 8.

### Solve the exercise Exercise 2 page 102 Math textbook 10 Kite episode 2

Let the ellipse $$\left( E \right)$$ have the canonical equation $$\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{ 25}} = 1$$. Find the coordinates of the intersections of $$\left( E \right)$$ with the Ox, Oy axes and the coordinates of the focal points of $$\left( E \right)$$.

Solution method

Ellipse (E) intersects 2 coordinate axes Ox, Oy at four points $${A_1}\left( { – a;{\rm{ }}0} \right)$$$${A_2}\left( { {) a{\rm{ }};{\rm{ }}0} \right)$$$${B_1}\left( {0; – {\rm{ }}b} \right)$$$${B_2 }\left( {0;{\rm{ }}b} \right)$$

Ellipse (E) has 2 focal points $${F_1}\left( { – c;0} \right)$$ and $${F_2}\left( {c;0} \right)$$ where \ ({a^2} = {c^2} + {b^2}\)

Solution guide

From the canonical equation of (E) we have: $$a = 7,b = 5 \Rightarrow c = 2\sqrt 6 {\rm{ }}(do{\rm{ }}{{\rm{c} }^2} + {b^2} = {a^2})$$

So we have the coordinates of the intersections of (E) with the axis Ox, Oy are: $${A_1}\left( { – 7;{\rm{ }}0} \right)$$$${A_2}\ left( {7;{\rm{ }}0} \right)$$$${B_1}\left( {0; – {\rm{ 5}}} \right)$$$${B_2}\left ( {0;{\rm{ 5}}} \right)$$

The two foci of (E) have coordinates: $${F_1}\left( { – 2\sqrt 6 ;0} \right),{F_2}\left( {2\sqrt 6 ;0} \right)$$

### Solve the exercise Exercise 3 page 102 Math textbook 10 Kite episode 2

Write the canonical equation of the ellipse $$\left( E \right)$$, knowing the coordinates of the two intersections of $$\left( E \right)$$ with Ox and Oy as $${A_1}\, respectively. left( { – 5;0} \right)$$ and $${B_2}\left( {0;\sqrt {10} } \right)$$

Solution method

Ellipse (E) intersects 2 coordinate axes Ox, Oy at four points $${A_1}\left( { – a;{\rm{ }}0} \right)$$$${A_2}\left( { {) a{\rm{ }};{\rm{ }}0} \right)$$$${B_1}\left( {0; – {\rm{ }}b} \right)$$$${B_2 }\left( {0;{\rm{ }}b} \right)$$

Solution guide

Since (E) intersects Ox at $${A_1}\left( { – 5;0} \right)$$ we have: $$a = 5$$

Since (E) intersects Oy at $${B_2}\left( {0;\sqrt {10} } \right)$$ we have: $$b = \sqrt {10}$$

So the canonical equation of (E) is: $$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{10}} = 1$$

### Solve exercises Lesson 4, page 102, Math textbook 10 Kite episode 2

We know that the Moon moves around the Earth in an ellipse of which the Earth is one focus. That ellipse has $${A_1}{A_2}$$ = 768 800 km and $${B_1}{B_2}{\rm{ }} = {\rm{ }}767{\rm{ }}619{\rm { }}km$$ (Source: Ron Larson (2014), Precalculus Real Mathematics, Real People, Cengage) (Figure 62). Write the canonical equation for that ellipse.

Solution method

The ellipse (E) has the canonical equation: $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^) 2}}} = 1\left( {a > b > 0} \right)$$, where: $${A_1}{A_2} = 2a,{B_1}{B_2} = 2b$$

Solution guide

We have:

$${A_1}{A_2} = 2a \Rightarrow 2a = 768800 \Rightarrow a = 384400$$ and $${B_1}{B_2} = 2b \Rightarrow 767619 = 2b \Rightarrow b = 383809.5$$

So the canonical equation of (E) is: $$\frac{{{x^2}}}{{{{384400}^2}}} + \frac{{{y^2}}}{{383809) ,5}} = 1$$

### Solve exercises Exercise 5 page 102 Math textbook 10 Kite episode 2

Which of the following equations are canonical equations of hyperbola?

a) $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{9} = 1$$ b) $$\frac{{{x^2} }}{9} – \frac{{{y^2}}}{9} = 1$$ c) $$\frac{{{x^2}}}{9} – \frac{{{y^) 2}}}{{64}} = 1$$ d) $$\frac{{{x^2}}}{{64}} – \frac{{{y^2}}}{9} = 1$$

Solution method

The hyperbola (H) has the canonical equation: $$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^) 2}}} = 1$$, where $${a^2} = {c^2} – {b^2}$$

Solution guide

The equations that are canonical equations of (H) are: b), c), d).

### Solve exercises Lesson 6 page 102 Math textbook 10 Kite episode 2

Find the coordinates of the focal points of the hyperbola in each of the following cases:

a) $$\frac{{{x^2}}}{9} – \frac{{{y^2}}}{{16}} = 1$$

b) $$\frac{{{x^2}}}{{36}} – \frac{{{y^2}}}{{25}} = 1$$

Solution method

The hyperbola (H) has the canonical equation: $$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^) 2}}} = 1$$, where $${a^2} = {c^2} – {b^2}$$

Solution guide

a) We have: $$a = 3,b = 4 \Rightarrow c = \sqrt {{3^2} + {4^2}} = 5$$

So the focus of (E) is: $${F_1}\left( { – 5;0} \right),{F_2}\left( {5;0} \right)$$

b) We have: $$a = 6;b = 5 \Rightarrow c = \sqrt {{6^2} + {5^2}} = \sqrt {61}$$

So the focus of (E) is: $${F_1}\left( { – \sqrt {61} ;0} \right),{F_2}\left( {\sqrt {61} ;0} \right)\ ) ### Solving exercises Exercise 7 page 102 Math textbook 10 Kite episode 2 Write the canonical equation of the hyperbol \(\left( H \right)$$, knowing that $$N\left( {\sqrt {10} ;2} \right)$$ lies on $$\left( H \right) )$$ and the coordinate of an intersection of $$\left( H \right)$$ with the Ox axis equals 3.

Solution method

The hyperbola (H) has the canonical equation: $$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^) 2}}} = 1$$, where $${a^2} = {c^2} – {b^2}$$

Hyperbola (H) intersects the Ox axis at two foci.

Solution guide

Since the hyperbola (H) intersects the Ox axis at a point with a coordinate of 3, we have: $${F_1}\left( {3;0} \right) \Rightarrow c = 3 \Rightarrow {a^2} + { b^2} = 9\left( 1 \right)$$

Since $$N\left( {\sqrt {10} ;2} \right) \in \left( H \right)$$ we have: $$\frac{{10}}{{{a^2} }} – \frac{4}{{{b^2}}} = 1\left( 2 \right)$$

From $$\left( 1 \right),\left( 2 \right)$$ we have: $$a = \sqrt 5 ,b = 2$$

So the canonical equation of (H) is: $$\frac{{{x^2}}}{5} – \frac{{{y^2}}}{4} = 1$$

### Solve exercises Exercise 8 page 102 Math textbook 10 Kite episode 2

Which of the following is the canonical equation of the parabola?

a) $${y^2} = – 2x$$

b) $${y^2} = 2x$$

c) $${x^2} = – 2y$$

d) $${y^2} = \sqrt 5 x$$

Solution method

The canonical equation of the parabola is: $${y^2} = 2px\left( {p > 0} \right)$$

Solution guide

The canonical equations of the parabola are: b), d)

### Solve the exercises Lesson 9 page 102 Math textbook 10 Kite episode 2

Find the coordinates of the focal point and write the equation of the standard curve of the parabola in each of the following cases:

a) $${y^2} = \frac{{5x}}{2}$$

b) $${y^2} = 2\sqrt 2 x$$

Solution method

The canonical equation of the parabola is: $${y^2} = 2px\left( {p > 0} \right)$$, where the focus is $$F\left( {\frac{p}{2 };0} \right)$$ and the standard curve equation is: $$x + \frac{p}{2} = 0$$.

Solution guide

a) The focal point of the parabola is: $$F\left( {\frac{5}{4};0} \right)$$

The standard curve equation is: $$x + \frac{5}{4} = 0$$

b) The focus of the parabola is: $$F\left( {\sqrt 2 ;0} \right)$$

The standard curve equation is: $$x + \sqrt 2 = 0$$

### Solution of Exercises Lesson 10, page 102, Math Textbook 10, Kite episode 2

Write the canonical equation of the parabola, knowing the focus $$F\left( {6;0} \right)$$

Solution method

The canonical equation of the parabola is: $${y^2} = 2px\left( {p > 0} \right)$$, where the focus is $$F\left( {\frac{p}{2 };0} \right)$$ and the standard curve equation is: $$x + \frac{p}{2} = 0$$.

Solution guide

Since the parabola has focus $$F\left( {6;0} \right)$$ we have $$\frac{p}{2} = 6 \Leftrightarrow p = 12$$

So the canonical equation of the parabola is: $${y^2} = 24x$$

### Solving exercises Lesson 11 page 102 Math textbook 10 Kite episode 2

A lamp whose cross-section is a parabola (Figure 63). The parabola has a width between the rims AB = 40 cm and a depth h = 30 cm (h is the distance from O to AB). The light bulb is at the focal point S. Write the canonical equation of the parabola.

Solution method

The canonical equation of the parabola is: $${y^2} = 2px\left( {p > 0} \right)$$, where the focus is $$F\left( {\frac{p}{2 };0} \right)$$ and the standard curve equation is: $$x + \frac{p}{2} = 0$$.

Solution guide

Let’s call the canonical equation of the parabola: $${y^2} = 2px\left( {p > 0} \right)$$

Since $$AB = 40cm$$ and $$h = 30cm$$ $$A\left( {30;20} \right)$$

Since $$A\left( {30;20} \right)$$ is parabolic, we have: $${20^2} = 2p.30 \Rightarrow p = \frac{{20}}{3}$$

So the parabola has the canonical equation: $${y^2} = \frac{{40}}{3}x$$