Question

In the parabola $y^2=4ax,$ , the locus of middle points of all chords of constant length C is

• A. $\left(4ax-y^2\right)\left(y^2-4a^2\right)=a^2{c}^2$
• B. $\left(4ax+y^2\right)\left(y^2+4a^2\right)=a^2{c}^2$
• C. $\left(4ax+y^2\right)\left(y^2-4a^2\right)=a^2{c}^2$
• D. $\left(4ax-y^2\right)\left(y^2+4a^2\right)=a^2{c}^2$

Answer 1

The correct option is D $\left(4ax-y^2\right)\left(y^2+4a^2\right)=a^2{c}^2$

Let $AB$ be the chord such that $\left|AB\right|=c$.

Let the coordinates of $A$ and $B$ are :

$A(a{t_1}^2,2a{t}_1)$ and $B(a{t_2}^2,2a{t}_2)$

If $P(h,k)$ be the middle point of $AB$, then

$(h,k)=(\frac{a{t_1}^2+a{t_2}^2}{2},\frac{2a{t}_1+2a{t}_2}{2})$

$\Rightarrow h=\frac{a{t_1}^2+a{t_2}^2}{2}$ and $k=\frac{2a{t}_1+2a{t}_2}{2}$

$\Rightarrow \frac{2h}{a}={t_1}^2+{t_2}^2$ and $\frac{k}{a}=t_1+t_2$

Now,$\frac{2h}{a}={t_1}^2+{t_2}^2$

$\Rightarrow \frac{2h}{a}={(t_1+t_2)}^2-2t_1{t}_2$

$\Rightarrow \frac{2h}{a}={\left(\frac{k}{a}\right)}^2-2t_1{t}_2$

$\Rightarrow 2t_1{t}_2=\frac{k^2}{a^2}-\frac{2h}{a}$

$\Rightarrow 2t_1{t}_2=\frac{k^2-2ah}{a^2}$

$\Rightarrow t_1{t}_2=\frac{k^2-2ah}{2a^2}$ and $t_1+t_2=\frac{k}{a}$

Since $\left|AB\right|=c$

$\Rightarrow {AB}^2=c^2$

$\Rightarrow {(a{t_1}^2-a{t_2}^2)}^2+{(2a{t}_1-2a{t}_2)}^2=c^2$

$\Rightarrow a^2{(t_1-t_2)}^2[{(t_1+t_2)}^2+4]=c^2$

$\Rightarrow a^2[{(t_1+t_2)}^2-4t_1{t}_2][{(t_1+t_2)}^2+4]=c^2$

$\Rightarrow a^2[\frac{k^2}{a^2}-\frac{4(k^2-2ah)}{a^2}][\frac{k^2}{a^2}+4]=c^2$

$\Rightarrow a^2\left[\frac{k^2-2k^2+4ah}{a^2}\right]\left[\frac{k^2+4a^2}{a^2}\right]=c^2$

$\Rightarrow (4ah-k^2)(k^2+4a^2)=a^2{c}^2$

Hence, locus of mid point of $AB$ is $(4ax-y^2)(y^2+4a^2)=a^2{c}^2$ by replacing $h\to x$ and $k\to y$