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Problem Solving Lesson 22: Three conic lines (Connecting)
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Solve problem 7.19 page 56 Math textbook 10 Connecting knowledge volume 2

Let the ellipse have the equation: \(\frac{x^{2}}{36}+\frac{y^{2}}{9}=1\). Find the focal point and focal length of the ellipse.

Solve problem 7.19 page 56 Math textbook 10 Connecting knowledge volume 2

Let the ellipse have the equation: \(\frac{x^{2}}{36}+\frac{y^{2}}{9}=1\). Find the focal point and focal length of the ellipse.

Solution method

The canonical equation of the ellipse has the form \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}} } = 1\) with \(a > b > 0\).

For a > b > 0, both are equations of an ellipse with two focal points \({F_1}\left( { – \sqrt {{a^2} – {b^2}} ;0} \right),{ F_2}\left( {\sqrt {{a^2} – {b^2}} ;0} \right)\), focal length \(2c = 2\sqrt {{a^2} – {b^2 }} \) and the sum of the distances from each point of the ellipse to the two foci is 2a.

Detailed explanation

I have: a² = 36, b² = 9, c = \(\sqrt{a^{2}-b^{2}}=\sqrt{27}\).

Focal F_first(\(-\sqrt{27}\);0) and F₂(\(\sqrt{27}\);0).

Focal length 2c = \(2\sqrt{27}\).

Solve lesson 7.20, page 56, Math textbook 10, Connecting knowledge volume 2

Let the hyperbola have the equation: \(\frac{x^{2}}{7}-\frac{y^{2}}{9}=1\). Find the focal point and focal length of the hyperbola.

Solve lesson 7.20, page 56, Math textbook 10, Connecting knowledge volume 2

Let the hyperbola have the equation: \(\frac{x^{2}}{7}-\frac{y^{2}}{9}=1\). Find the focal point and focal length of the hyperbola.

Solution method

The canonical equation of the hyperbola has the form \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}} } = 1\) with \(a,b > 0\).

For a, b > 0, both are equations of hyperbola with two foci \({F_1}\left( { – \sqrt {{a^2} + {b^2}} ;0} \right),{ F_2}\left( {\sqrt {{a^2} + {b^2}} ;0} \right)\), focal length \(2c = 2\sqrt {{a^2} + {b^2 }} \) and the absolute value of the difference of the distances from each hyperbolic point to the two foci is 2a.

Detailed explanation

I have: a² = 7, b² = 9, c = \(\sqrt{a^{2}+b^{2}}=4\).

Focal F_first(-4;0) and F₂(4;0).

Focal length 2c = 8

Solve problem 7.21 page 56 Math textbook 10 Connecting knowledge volume 2

Let the parabola have the equation: \({y^2} = 8x\). Find the focus and standard curve of the parabola.

Solve problem 7.21 page 56 Math textbook 10 Connecting knowledge volume 2

Let the parabola have the equation: \({y^2} = 8x\). Find the focus and standard curve of the parabola.

Solution method

The canonical equation of the parabola has the form \({y^2} = 2p{\rm{x}}\) (with p > 0)

For p > 0, is the canonical equation of the parabola with focus \(F\left( {\frac{p}{2};0} \right)\) and the standard curve \(\Delta 😡 = – \frac {p}{2}\).

Detailed explanation

We have: 2p = 8 so p = 4.

Focal point F(2; 0) and datum \(\Delta \): x = -2.

Solve problem 7.22 page 56 Math textbook 10 Connecting knowledge volume 2

Draw the canonical equation of the ellipse that passes through the point A(5; 0) and has a focal point F2(3; 0).

Solve problem 7.22 page 56 Math textbook 10 Connecting knowledge volume 2

Draw the canonical equation of the ellipse that passes through the point A(5; 0) and has a focal point F2(3; 0).

Solution method

The ellipse (E) has the form: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

+ (E) goes through A(5; 0) and replaces the A coordinate into (E) => a

+ (E) has focal point F₂(3; 0) => b

Detailed explanation

The ellipse (E) has the form: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with a>b >0.

+ (E) goes through A(5; 0) so \(\frac{5^{2}}{a^{2}}+\frac{0^{2}}{b^{2}}=1 \)

=> a = 5.

+ (E) has focal point F₂(3; 0) so c = 3

=> b = \(\sqrt{a^{2}-c^{2}}=4\)

So the canonical equation of (E): \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Solve problem 7.23 page 56 Math textbook 10 Connecting knowledge volume 2

Write the canonical equation of the parabola passing through the point M(2; 4).

Solve problem 7.23 page 56 Math textbook 10 Connecting knowledge volume 2

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Write down the canonical equation of the parabola passing through the point M(2; 4).

Solution method

The canonical equation of the parabola has the form \({y^2} = 2p{\rm{x}}\) (with p > 0)

For p > 0, is the canonical equation of the parabola with focus \(F\left( {\frac{p}{2};0} \right)\) and the standard curve \(\Delta 😡 = – \frac {p}{2}\).

Detailed explanation

The parabolic equation (P) has the form: y² = 2px.

(P) goes through M(2; 4) so ​​4² = 2p.2

=> 2p = 8

So equation (P): y² = 8x.

Solution 7.24 page 56 Math textbook 10 Connecting knowledge volume 2

There are two radio signal stations located at two locations A and B, 300 km apart. At the same time, the two stations simultaneously transmit a signal with a speed of 292 000 km/s for a ship to receive and measure the time difference. The signal from A arrives 0.0005 s earlier than the signal from B. From the above information, we can determine which hypobola the ship belongs to? Write down the canonical equation for that hyperbola in kilometers.

Solution 7.24 page 56 Math textbook 10 Connecting knowledge volume 2

There are two radio signal stations located at two locations A and B, 300 km apart. At the same time, the two stations simultaneously transmit a signal with a speed of 292 000 km/s for a ship to receive and measure the time difference. The signal from A arrives 0.0005 s earlier than the signal from B. From the above information, we can determine which hypobola the ship belongs to? Write down the canonical equation for that hyperbola in kilometers.

Solution method

The canonical equation of the hyperbola has the form \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}} } = 1\) with \(a,b > 0\).

For a, b > 0, both are equations of hyperbola with two foci \({F_1}\left( { – \sqrt {{a^2} + {b^2}} ;0} \right),{ F_2}\left( {\sqrt {{a^2} + {b^2}} ;0} \right)\), focal length \(2c = 2\sqrt {{a^2} + {b^2 }} \) and the absolute value of the difference of the distances from each hyperbolic point to the two foci is 2a.

Detailed explanation

Choose the coordinate system Oxy so that A, B lie on the axis Ox, the ray Ox coincides with the ray OB, O is the midpoint of AB. So the coordinates of two points are: A(-150; 0) and B(150; 0)

Then the position of the ship is the point M lying on the hyperbola with 2 focal points A and B.

The signal from A arrives earlier than the signal from B by 0.0005 s, so we have: |MA – MB| = 0.0005.292 000 = 146 km.

Let the canonical equation of the hyperbola take the form: \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) with a, b > 0.

Do |MA – MB| = 146 = 2a <=> a = 73.

Since the two foci are: A(-150; 0) and B(150; 0) so c = 150

=> b = \(\sqrt{c^{2}-a^{2}}=\sqrt{17171}\)

So the canonical equation of the hyperbola to find is: \(\frac{x^{2}}{5329}-\frac{y^{2}}{17171}=1\)

Solve problem 7.25, page 56, Math textbook 10, Connecting knowledge volume 2

The bend of a road is in the shape of a parabola, the entry point of the bend is A, the end point is B, the distance AB = 400m. The parabolic vertex (P) of the segment is about 20 m from line AB and equidistant from A and B.

a) Make the canonical equation of (P), with 1 unit of measure in the coordinate plane corresponding to 1 m in reality.

b) Make the canonical equation of (P), with 1 unit of measure in the coordinate plane corresponding to 1 km in fact.

Solve problem 7.25, page 56, Math textbook 10, Connecting knowledge volume 2

The bend of a road is in the shape of a parabola, the entry point of the bend is A, the end point is B, the distance AB = 400m. The parabolic vertex (P) of the segment is about 20 m from line AB and equidistant from A and B.

a) Make the canonical equation of (P), with 1 unit of measure in the coordinate plane corresponding to 1 m in reality.

b) Make the canonical equation of (P), with 1 unit of measure in the coordinate plane corresponding to 1 km in fact.

Solution method

The canonical equation of the parabola has the form \({y^2} = 2p{\rm{x}}\) (with p > 0)

For p > 0, is the canonical equation of the parabola with focus \(F\left( {\frac{p}{2};0} \right)\) and the standard curve \(\Delta 😡 = – \frac {p}{2}\).

Detailed explanation

Choose the coordinate system so that the vertex of the parabola coincides with the origin O(0; 0) (as shown).

a) If a unit of measurement in the coordinate plane corresponds to 1 m in reality, the coordinates of the points are: A(20; -200) and B(20; 200) belonging to a parabola of the form y² = 2px

Substitute the coordinates of point A and we have: 200² = 2p.20 => 2p = 2000

So the parabola has the form: y² = 2000.x

b) If 1 unit of measurement in the coordinate plane corresponds to 1 km in reality, the coordinates of the points are: A(0.02; -0.2) and B(0.02; 0.2) belong to parabola of the form y² = 2px

Substitute the coordinates of point A and we have: 0.2² = 2p.0.02 => 2p = 2

So the parabola has the form: y² = 2.x