Locus of the midpoint of any normal chords of $y^{2} = 4 a x$ is

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = a (\frac{4 a^{2}}{y^{2}} + 2 + \frac{y^{2}}{2 a^{2}})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = a (\frac{4 a^{2}}{y^{2}} - 2 - \frac{y^{2}}{2 a^{2}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x = a (\frac{4 a^{2}}{y^{2}} + 2 + \frac{y^{2}}{2 a^{2}})$
Let $A B$ be a normal chord where $A \equiv (a t_{1}^{2}, 2 a t_{1})$ and $B \equiv (a t_{2}^{2}, 2 a t_{2})$ .
Let the midpoint of $A B$ is $P (h, k)$ , then
$2 h = a (t_{1}^{2} + t_{2}^{2})$
$= a [(t_{1} + t_{2})^{2} - 2 t_{1} t_{2}]$
and $2 k = 2 a (t_{1} + t_{2})$
We also have,
$t_{2} = - t_{1} - \frac{2}{t_{1}}$
$\Rightarrow t_{1} + t_{2} = \frac{- 2}{t_{1}}$ and $t_{1} t_{2} = - t_{1}^{2} - 2$
$\Rightarrow \frac{k}{a} = \frac{- 2}{t_{1}}$
$\Rightarrow t_{1} = - \frac{2 a}{k}$
So, $2 h = a [(- \frac{2}{t_{1}})^{2} + 2 t_{1}^{2} + 4]$
$\Rightarrow h = a (\frac{2}{t_{1}^{2}} + t_{1}^{2} + 2)$
Thus, required locus is $x = a (\frac{4 a^{2}}{y^{2}} + 2 + \frac{y^{2}}{2 a^{2}})$

Suggest Corrections

Similar questions

2

x^{2} + y^{2} = 2 a^{2}

y^{2} = 8 a x

x^{2} + y^{2} = \sqrt{2} a^{2}

x^{2} - y^{2} = a^{2}

The equation of the common tangent to $x^{2} + y^{2} = 2 a^{2}$ and $y^{2} = 8 a x$ is

List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^{2} + y^{2} = a^{2}$	1) $x^{2} + y^{2} = 5$ of
B. The locus of the mid-point of the chord of the circle $x^{2} + y^{2} = 8$ which is $\sqrt{2}$ units from origin is	2) $x^{2} + y^{2} = a^{2} {sec}^{2} α$
C. The locus of the point of intersection of the lines $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$	3) $x^{2} + y^{2} = 2 a^{2}$
D. The locus of the point whose chord of contact w.r.t $x^{2} + y^{2} = a^{2}$ making an angle '2 $α$ ' at the center is	4) $x^{2} + y^{2} = 2$ 5) $x^{2} + y^{2} = 1$

(x - 2 a)^{2} + y^{2} = x^{2}

Join BYJU'S Learning Program