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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
adsense
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: \({y^2} = \frac{{40}}{3}x\)
