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The line 2x+5y=1 meets the curve x^2+5y-4y^2+10=0 at the point A and B.

Sagot :

abierasrochellegay

Answer:

dont know sorryyyyyyyyyy

haytambelhaj

Answer:

A(3,-1) and B( -2,1)

Step-by-step explanation:

2x+5y=1. , or. y=(1–2x)/5………….(1)

x^2+5xy-4y^2+10=0…………………..(2)

Putting y=(1–2x)/5 from eqn. (1)

x^2+5x.(1–2x)/5–4/25.(1–2x)^2 +10=0

or. 25x^2+25x.(1–2x)-4.(1+4x^2–4x)+250 = 0

or. 25x^2+25x-50x^2–4–16x^2+16x+250=0

or. 41x^2–41x-246=0

or x^2– x - 6=0

or. (x-3).(x+2)=0

x= 3 or -2

But y=(1–2x)/5

y= (1–6)/5 or. (1+4)/5

y= -1. or. 1

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