ParabolaHardBloom L3

Question

The locus of the middle points of chords of the parabola $y^{2} = 4x$, which are of constant length ' $2l$ ' is :

Options

A.$\left( 4x + y^{2} \right)\left( y^{2} - 4 \right) = 4l^{2}$
B.$\left( 4y + x^{2} \right)\left( x^{2} - 4 \right) = 4l^{2}$
C.$\left( 4y - x^{2} \right)\left( x^{2} + 4 \right) = 4l^{2}$
D.$\left( 4x - y^{2} \right)\left( y^{2} + 4 \right) = 4l^{2}$

Solution

Sol. [Key Concept: Use the midpoint-chord relation $T = S_{1}$ to express the chord's direction and length through its midpoint, then impose length $2l$.]

Put $(h + rcos\theta,k + rsin\theta)$ to $y^{2} = 4x$

$$\begin{matrix} & \ \Rightarrow k^{2} + r^{2}\sin^{2}\theta + 2krsin\theta = 4h + 4rcos\theta \\ & \ \Rightarrow r^{2}\sin^{2}\theta + (2ksin\theta - 4cos\theta)r \\ & \ + \left( k^{2} - 4h \right) = 0^{- l} \\ & \ l + ( - l) = 2ksin\theta - 4cos\theta = 0 \\ & \ \Rightarrow \ tan\theta = \frac{2}{k} \end{matrix}$$

$${l( - l) = \frac{k^{2} - 4h}{\sin^{2}\theta} = \left( k^{2} - 4h \right)\left( 1 + \cot^{2}\theta \right) }{\Rightarrow \ \left( 4x - y^{2} \right)\left( 1 + \frac{y^{2}}{4} \right) = l^{2} }{\Rightarrow \left( 4x - y^{2} \right)\left( 4 + y^{2} \right) = 4l^{2} }$$

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