Question

设抛物线y2=4x与直线y=2x+k相交所得弦长为|AB|=3根号5.（1）求k

Answer 1

将直线y=2x+k带入y^2=4x,

∴4x^2+(4k-4)x+k^2=0

设两点的横坐标是x1,x2

相应的纵坐标为2x1+k,2x2+k

∵│AB│=3√5,

∴3√5=√[(x1-x2)^2+(y1-y2)^2]

=√((x1-x2)^2+(2x1-2x2)^2)

=√5(x1-x2)^2

=√(5(x1+x2)^2-20x1x2)

∵x1+x2=k-1,

又∵x1x2=k^2/4

∴3√5=√(5*(k-1)^2-20*k^2/4)

∴k=-4