Question

A point $P$ moves such that sum of the slopes of the normals drawn from it to the hyperbola $xy=16$ is equal to the sum of ordinates of feet of normals. The locus of $P$ is a curve $C$.

The equation of the curve $C$ is :

• A. $x^2=4y$
• B. $x^2=16y$
• C. $x^2=12y$
• D. $y^2=8x$

Answer 1

The correct option is B $x^2=16y$

Given equation of hyperbola is $xy=16$

Let $(4t,\frac{4}{t})$ be any point on the hyperbola.
Slope of tangent at $(4t,\frac{4}{t})=-\frac{1}{t^2}$
Equation of normal at this point is

$y-\frac{4}{t}=t^2(x-4t)$
If the normal passes through P(h, k), then $k-\frac{4}{t}=t^2(h-4t)$

$\Rightarrow 4t^4-t^{3}h+tk-4=0$
This equation has roots $t_1,t_2,t_3,t_4$ which are parameters of the feet of four normals on the hyperbola. Therefore,

$\sum t_1=\frac{h}{4}$
$\sum t_1{t}_2=0$

$\sum t_1{t}_2{t}_3=-\frac{k}{4}$ .....(i)
$t_1{t}_2{t}_3{t}_4=-1$ .....(ii)
Dividing (i) by (ii), we get

$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}=\frac{k}{4}$

$\Rightarrow \frac{4}{t_1}+\frac{4}{t_2}+\frac{4}{t_3}+\frac{4}{t_4}=k$

$\Rightarrow y_1+y_2+y_3+y_4=k$
According to the question,
$t_1^2+t_2^2+t_3^2+t_4^2=y_1+y_2+y_3+y_4$

$\because t_1^2+t_2^2+t_3^2+t_4^2=(t_1+t_2+t_3+t_4{)}^2-2(t_1{t}_2+t_1{t}_3+t_1{t}_4+t_2{t}_3+t_2{t}_4+t_3{t}_4)$

$\Rightarrow \frac{h^2}{16}=k$

Hence, the locus of (h, k) is

$x^2=16y$