Question

The chord $x+y=1$ of the curve $y^2=12x$ cuts it at the points $A$ and $B.$ The normals at $A$ and $B$ intersect at $C.$ If a third line from $C$ cuts the curve normally at $D,$ then the co-ordinates of $D$ are

• A. $(3,-6)$
• B. $(12,-12)$
• C. $(6,-3)$
• D. $(12,12)$

Answer 1

The correct option is D $(12,12)$

Let the ordinates of conormal points $A,B,D$ be $y_1,y_2,y_3$ respectively.
$\Rightarrow y_1+y_2+y_3=0$
$A$ and $B$ lie on line $x+y=1$
On solving, $x+y=1$ and $y^2=12x$
$y^2=12(1-y)$
$\Rightarrow y^2+12y-12=0$
$\Rightarrow y_1+y_2=-12$
Since, $y_1+y_2+y_3=0,$
$\Rightarrow y_3=12$
$\therefore x_3=\frac{y_3^2}{12}=12$
Hence, coordinates of $D$ are $(12,12)$