Question

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

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

Answer 1

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

Let $|AB|=c$ be the chord

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

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

$h=\frac{a{t}_1^2+a{t}_2^2}{2}$

$k=\frac{2a{t}_1+2a{t}_2}{2}$

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

$\Rightarrow \frac{k}{a}=t_1+t_2$

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

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

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

$\Rightarrow t_1+t_2=\frac{k}{a}$

Since, $|AB|=c$

$AB=c^2$

$(a{t}_1^2-a{t}_2^2{)}^2+(2a{t}_1-2a{t}_2{)}^2=c^2$

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

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

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

upon further solving the above equation, we get,

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

Hence the locus of midpoint is

$(4ax-y^2)(y^2+4a^2)=a^2{c}^2$