Find the equation of the straight line joining the origin to the point of intersection of the line 3x+4y-5=0 and the curve 2xsquare+3ysq =5
Answer (1)
30th Nov, 2019
3x+4y=5..(i)
2x^2+3y^2=5...(ii)
put the value of y form equation i to the equation ii
And we get
2x^2+3((5-3x)/4)^2=5
or, 2x^2+3/16(25+9x^2-30x)=5
or,32x^2+75+27x^2-90x=80
or,59x^2-90x-5=0
Hence the solution of x are
45/59-4/59*145
45/59+4/59*145
Take one of them as x1 and y1=(5-3x1)/5
Hence the line is y=y1/x1*(x)
2x^2+3y^2=5...(ii)
put the value of y form equation i to the equation ii
And we get
2x^2+3((5-3x)/4)^2=5
or, 2x^2+3/16(25+9x^2-30x)=5
or,32x^2+75+27x^2-90x=80
or,59x^2-90x-5=0
Hence the solution of x are
45/59-4/59*145
45/59+4/59*145
Take one of them as x1 and y1=(5-3x1)/5
Hence the line is y=y1/x1*(x)
Comments (0)
