Let a tangent to the curve y^2=24x meet the curve xy=2 at the points A and B.

Question

Let a tangent to the curve $y^2=24x$ meet the curve $xy=2$ at the points $A$ and $B$. Then the midpoints of such line segments $AB$ lie on a parabola with:

Accepted Answer

Given parabola ${y^2} = 24x = 4.6x = 4ax$

$$\therefore$$ a = 6

If $y = mx + c$ is a tangent to the parabola then,

$$c = {a \over m}$$

$$ \Rightarrow c = {6 \over m}$$

Let midpoint of chord AB is $({x_1},{y_1})$

We know, locus of midpoint of chord for hyperbola,

$T = {S_1}$

$$ \Rightarrow {{x{y_1} + {x_1}y} \over 2} - 2 = {x_1}{y_1} - 2$$

$$ \Rightarrow x{y_1} + {x_1}y = 2{x_1}{y_1}$$

$$ \Rightarrow y = \left( { - {{{y_1}} \over {{x_1}}}} \right)x + 2{y_1}$$

This equation is the same tangent to the parabola.

Then comparing with $y = mx + c$ we get,

$$c = 2{y_1},\,m =  - {{{y_1}} \over {{x_1}}}$$

$$\therefore$$ $$2{y_1} = {6 \over { - {{{y_1}} \over {{x_1}}}}}$$

$$ \Rightarrow  - y_1^2 = 3{x_1}$$

$$\therefore$$ Locus of midpoint $({x_1},{y_1})$ is,

$- {y^2} = 3x$

$$ \therefore $$ Length of latus rectum = 3

Equation of directrix is $x=\frac{3}{4}$

Solutions · 1