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3 years ago

Select the conic section that represents the equation. y2 = 10x circle ellipse parabola hyperbola

Mathematics

1 answer:

Makovka662 [10]3 years ago

6 0

The conic that represents the equation y² = 10x is a parabola

Step-by-step explanation:

The general equation for any conic section is

Ax² + Bxy + Cy² + Dx + Ey + F = 0, where A, B, C, D, E and F are

constants

1. If B² - 4AC < 0, if a conic exists, it will be either a circle or an ellipse,

if A and C are equal then it is a circle, if not then it is an ellipse

2. If B² - 4AC = 0, if a conic exists, it will be a parabola

3. If B² - 4AC > 0, if a conic exists, it will be a hyperbola

∵ The equation is y² = 10x

- Subtract 10x from both sides

∴ y² - 10x = 0

∴ A = 0 , B = 0 , C = 1 , D = -10 , E = 0 and F = 0

∵ B² - 4AC = (0)² - 4(0)(1) = 0

∴ B² - 4AC = 0

∴ The conic is parabola

The conic that represents the equation y² = 10x is a parabola

Learn more:

You can learn more about discriminant in brainly.com/question/8196933

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<h2>○=> <u>Correct options</u> :</h2><h2>□ $\color{plum}\tt\bold{(C) \: SR = 27}$ </h2><h2>□ $\color{plum}\tt\bold{(D) \: QR = 54 }$ </h2><h3><u>Steps to derive the correct options</u> :</h3>

Since two sides and one included angle is equal in △PQS and △PRS, we can conclude that they are congruent under the SAS congruence criterion.

Which means :

▪︎Angle S = Angle S

▪︎PS = PS

▪︎QS = RS

Given :

Measure of segment QS = 6n+3

Measure of segment RS = 4n+11

Thus :

$= \tt6n + 3 = 4n + 11$

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$=\tt n = \frac{8}{2}$

$\hookrightarrow\color{plum}\tt n = 4$

Thus, the value of n = 4

Measure of segment QS :

$=\tt 6n + 3$

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$\color{plum}\tt \: \bold{QS = 6n + 3 = \tt\bold{27}}$

Thus, measure of QS = 27

Measure of RS :

$=\tt 4n + 11$

$=\tt 4 × 4 + 11$

$=\tt 16+11$

$\color{plum}\tt \: \bold{RS = 4n + 11 = \tt\bold{27}}$

Measure of QR :

$=\tt 27+27$

$\color{plum}\tt \: \bold{QR = 27+27 \tt\bold{=54}}$

Thus :

▪︎QS = 27

▪︎RS = 27

▪︎QR = 54

Therefore, the correct options are :

▪︎(C) SR = 27

▪︎(D) QR = 54

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