Question

Help me guys!

Show that the locus of the middle points of normal chords of the parabola y² = 4ax is
y⁴ - 2a(x - 2a)y² + 8a⁴=0​

Answer 1

Chord of Contact:-

• The chord joining the points of contact of two tangents drawn from an external point to a parabola is known as the chord of contact of tangents drawn from external point.

Equation of the normal chord at any point (at², 2at) of the parabola y² = 4ax is

y + tx = 2at + at³ ....(i)

Look at the attached figure

But if M (x₁, y₁) be it's middle point its equation must be also,

T = S₁

:⟹ yy₁ - 2a (x + x₁) = y₁² - 4ax₁

:⟹ yy₁ - 2ax = y₁² - 2ax₁ .....(ii)

Therefore, from eqs. (i) and (ii) are identical, comparing, them

`\sf (1)/(y_1) = (t)/( - 2a) = \frac{2at + {at}^(3) }{ {y_1}^(2) - 2ax} \\ \\ \sf \: from \: first \: two \: relations \: ,t = - (2a)/(y_1) ....(iii)\\ \\ \sf \: from \: first \: two \: relations \: , \: (t)/( - 2a) = \frac{2at + {at}^(3) }{{y_1}^(2) - 2ax_1}`

`\implies \sf \: \frac{{y_1}^(2) - 2ax_1}{ - 2a} = 2a + {at}^(2) \: \\ \\ \implies \sf \: \frac{{y_1}^(2) - 2ax_1}{ - 2a} = 2a + a \bigg \lgroup ( - 2a)/(y_1) { \bigg \rgroup}^(2) \qquad\{ \: from \: eqs. \: (iii) \} \\ \\ \implies \sf \:\frac{{y_1}^(2) - 2ax_1}{ - 2a} = \frac{2a{y_1}^(2) + 4 {a}^(3) }{{y_1}^(2)} \\ \\ \implies \sf \: {y_1}^(4) - 2ax_1 {y_1}^(2) = - 4 {a}^(2) {y_1}^(2) - 8 {a}^(4) \\ \\ \implies \sf \: {y_1}^(4) - 2a(x_1 - 2a){y_1}^(2) + 8 {a}^(4) = 0`

`\sf hence, \: the \: locus \: of \: middle \: point \: (x_1,y_1) \: is \: \\ \\ \qquad \qquad \: \sf \: {y_1}^(4) - 2a(x_1 - 2a){y_1}^(2) + 8 {a}^(4) = 0`