Question

Prove that , the normal to $y^2=12x$ at (3,6) meets the parabola again in $(27,-18)$ & circle on this normal chord as diameter is $x^2+y^2-30x+12y-27=0$

Answer 1

$y^2=12x=4\times 3\times x$

$\therefore a=3$

point $(3,6)$ compare it withs $(a{t}^2,2at)$

we get $6=2at$

$\therefore t=\frac{6}{2a}=\frac{6}{2\times 3}=1$

$\therefore$ equation of normal

$y+tx=2at+a{t}^3$

$y+x=6+3$

$y=9-x$

on this normal intersects again as parabola

$\therefore y^2=12x$

or $(9-x{)}^2=12x$

or $81-18x+x^2=12x$.

$x^2-30x+81=0$

$x^2-27x-3x+81=0$

$x(x-27)-3(x-27)=0$

$(x-3)(x-27)=0$

$\therefore x=3$ or $27$

$x=3$ is already taken as is point of normal

$\therefore x=27$ then $y=9-x=9-27=-18$

Hence parabola meets normal at $(27,-18)$

for required circle

Centre $=(\frac{3+27}{2},\frac{6-18}{2})=(15,-6)$

Radius $=\sqrt{(15-3{)}^2+(-6-6{)}^2}$

$=\sqrt{{12}^2+{12}^2}=12\sqrt{2}$

$\therefore$ Equation of circle

$(x-15{)}^2+\{y-(-6){\}}^2=(12\sqrt{2}{)}^2$

$x^2-30x+225+y^2+12y+36=288$

or $x^2+y^2-30x+12y-27=0$

Hence proved.