Question

Help me guys!

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

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

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

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