Jadon Melendez

2022-07-27

The line y = mx + c intersects theparabola ${y}^{2}=4ax$ at the points P and Q.Show that the coordinates of the mid-point of PQ is $(\frac{2a-mc}{{m}^{2}},\frac{2a}{m})$.

If the mid-point is M, find the locus of M whenm varies and c = 1

If the mid-point is M, find the locus of M whenm varies and c = 1

losnonamern

Beginner2022-07-28Added 12 answers

y=mx+c eqn (1)

${y}^{2}=ax$ eqn (2)

Square both side eqn 1

will get

${y}^{2}={m}^{2}{x}^{2}+2mcx+{c}^{2}$... (3)

set 2 and 3 equals

$4ax={m}^{2}{x}^{2}+2mcx+{c}^{2}$

or

$({m}^{2}){x}^{2}+(2mc-4a)x+({c}^{2})=0$

$x=-\frac{b}{2a}=\frac{-(2mc-4a)}{2{m}^{2}}=\frac{2a-mc}{{m}^{2}}$

now plug in function 1

$y=m\ast (\frac{2a-mc}{{m}^{2}})+c=\frac{2a-mc}{m}+\frac{cm}{m}=\frac{2a}{m}$

so we just showed that it works

$(\frac{2a-mc}{{m}^{2}},\frac{2a}{m})$

next,

when c=1:

$(\frac{2a-m}{{m}^{2}},\frac{2a}{m})$

${y}^{2}=ax$ eqn (2)

Square both side eqn 1

will get

${y}^{2}={m}^{2}{x}^{2}+2mcx+{c}^{2}$... (3)

set 2 and 3 equals

$4ax={m}^{2}{x}^{2}+2mcx+{c}^{2}$

or

$({m}^{2}){x}^{2}+(2mc-4a)x+({c}^{2})=0$

$x=-\frac{b}{2a}=\frac{-(2mc-4a)}{2{m}^{2}}=\frac{2a-mc}{{m}^{2}}$

now plug in function 1

$y=m\ast (\frac{2a-mc}{{m}^{2}})+c=\frac{2a-mc}{m}+\frac{cm}{m}=\frac{2a}{m}$

so we just showed that it works

$(\frac{2a-mc}{{m}^{2}},\frac{2a}{m})$

next,

when c=1:

$(\frac{2a-m}{{m}^{2}},\frac{2a}{m})$

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