The locus of the middle points of the normàl chords of the parabola, y2=4ax is

Question

The locus of the middle points of the norm&agrave;l chords of the parabola, y2=4ax is (A) x+2a=(y2/2 a)+(4 a3/y2) (B) x+2a=(y2/2 a)-(4 a3/y2) (C) x-2a=(y2/2 a)+(4 a3/y2) (D) None of these

Tardigrade Admin · Accepted Answer

Here, equation of the normal chord at any point

a t^{2}, 2 a t

) of the parabola is

y + t x = 2 a t + a t^{3} \dots

(i)
Equation of the chord with mid-point

x_{1}, y_{1}

is

T = S_{1}

y y_{1} - 2 a x + x_{1} = y_{1}^{2} - 4 a x_{1}

y y_{1} - 2 a x = y_{1}^{2} - 2 a x_{1} \dots . (ii)

since, Eqs. (i) and (ii) are identical

\frac{1}{y _{1}} = \frac{t}{- 2 a} = \frac{2 a t + a t ^{3}}{y _{1}^{2} - 2 a x _{1}}

t = \frac{- 2 a}{y _{1}}

and

\frac{y _{1}^{2} - 2 a x _{1}}{- 2 a} = \frac{2 a t + a t ^{3}}{t}

= 2 a + a \frac{- 2 a ^{2}}{y _{1}}

or \frac{- y _{1}^{2}}{2 a} + x_{1} = 2 a + \frac{4 a ^{3}}{y _{1}^{2}}

\Rightarrow x_{1} - 2 a = \frac{y _{1}^{2}}{2 a} + \frac{4 a ^{3}}{y _{1}^{2}}

Hence, the locus of the middle point

x_{1}, y_{1}

is

x - 2 a = \frac{y ^{2}}{2 a} + \frac{4 a ^{3}}{y ^{2}}

Q. The locus of the middle points of the normàl chords of the parabola, $y^{2} = 4 a x$ is

$x + 2 a = \frac{y ^{2}}{2 a} + \frac{4 a ^{3}}{y ^{2}}$

$x + 2 a = \frac{y ^{2}}{2 a} - \frac{4 a ^{3}}{y ^{2}}$

$x - 2 a = \frac{y ^{2}}{2 a} + \frac{4 a ^{3}}{y ^{2}}$

None of these

Q. The locus of the middle points of the normàl chords of the parabola, y2=4ax is

x+2a=2ay2​+y24a3​

x+2a=2ay2​−y24a3​

x−2a=2ay2​+y24a3​

None of these

Q. The locus of the middle points of the normàl chords of the parabola, $y^{2} = 4 a x$ is

$x + 2 a = \frac{y ^{2}}{2 a} + \frac{4 a ^{3}}{y ^{2}}$

$x + 2 a = \frac{y ^{2}}{2 a} - \frac{4 a ^{3}}{y ^{2}}$

$x - 2 a = \frac{y ^{2}}{2 a} + \frac{4 a ^{3}}{y ^{2}}$