Question

Locus of mid points of the chords of the parabola $y^2=4ax$ which touch the circle $x^2+y^2=a^2$ is

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

Answer 1

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

Given parabola $y^2=4ax$
Let $M(x_1,y_1)$ be the mid-point of any chord of the parabola.
Then its equation is given by $T=S_1$
$\Rightarrow y{y}_1-2a(x+x_1)=y_1^2-4a{x}_1$
$\Rightarrow 2ax-y{y}_1+y_1^2-2a{x}_1$ ....(1)
Since, it touches the circle $x^2+y^2=a^2$
So, length of perpendicular from $(0,0)$ to (1) $=$ radius $a$
$\frac{|0-0+y_1^2-2a{x}_1|}{\sqrt{4a^2+y_1^2}}=a$
$\Rightarrow (y_1^2-2a{x}_1{)}^2=a^2(4a^2+y_1^2)$
Hence, locus of midpoint of chords is $(y^2-2ax{)}^2=a^2(4a^2+y^2)$