Question

x^2+3y=3; x^2-y^2+25=0​

Answer 1

Answer:

x

Step-by-step explanation:

Given : curves

x² + 3y  = 3

x² - y²  + 25 = 0

To Find : angle between the curves

Solution:

x² + 3y  = 3

x² - y²  + 25 = 0 => x² = y² - 25

y² - 25   + 3y  = 3

=> y²    + 3y  - 28  = 0

=> y² + 7y - 4y - 28 = 0

=> (y + 7)(y - 4) = 0

=> y = -7  ,  4

x² = y² - 25  => x² = 24  or  - 9

- 9 not possible

Hence x² = 24

=> x = ±2√6

Points of intersection   ±2√6 , -7

Taking  point

2√6 , -7

x² + 3y  = 3  =>    dy/dx  = -2x/3    = -4√6/3

x² - y²  + 25 = 0 =>  dy/dx  =  x/y   = -2√6 /7

Angle between line = α

tan  α = | (-4√6/3 - (-2√6 /7))/(1 + (-4√6/3)( -2√6 /7)|

=>  tan  α =| (-28√6 +6√6) (21 + 48) |

=> tan  α = | (-22√6) /69 |

=>  tan  α = | (-22√6) /69 |

α = 37.99°

angle between curves = 38°     or  152°

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