Question

A point P moves such that sum of the slopes of the normal drawn from it to the hyperbola $xy=16$ is equal to the sum of the ordinates of the feet of the normal. Let 'P' lies on the curve C, then.
The equation of '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 $x^2=16y$

Let P(h,k) be the point in the plane of hyperbola $xy=16=4^2$

The equation of the normal at a point $(4t,\frac{4}{t})$ to the hyperbola $xy=4^2$ is

$x{t}^3-yt-4t^4+4=0$

if if passes through $P(h,k)$ then,

$h{t}^3-yt-4t^4+4=0$

$\Rightarrow 4t^4-h{t}^3+kt-4=0$

this is a fourth degree equation in t. So, it gives 4 values of t, $t_1,t_2,t_3,t_4$ corresponding to each value of t there is a point on hyperbola such that the normal passes through $P(h,k)$.

let the 4 points be $A(c{t}_1,\frac{c}{t_1})$,$B(c{t}_2,\frac{c}{t_2})$,$C(c{t}_3,\frac{c}{t_3})$ and $D(c{t}_4,\frac{c}{t_4})$

such that normal at these points pass through $P(h,k)$.

Since $t_1,t_2,t_3,t_4$ are roots of equation (1).

Therefore,

$t_1+t_2+t_3+t_4=\frac{h}{c}=\frac{h}{4}$.......(2)

$\sum t_1{t}_2=0$.......(3)

$\sum t_1{t}_2{t}_3=-\frac{k}{4}$.........(4)

$t_1{t}_2{t}_3{t}_4=-1$.......(5)

It s given that the sum of the slopes of the normals at A,B,C and D is equal to the sum of the coordinates of these points.

Therefore

$t_1^2+t_2^2+t_3^2+t_4^2=\frac{c}{t_1}+\frac{c}{t_3}+\frac{c}{t_3}+\frac{c}{t_4}=\frac{4}{t_1}+\frac{4}{t_2}+\frac{4}{t_3}+\frac{4}{t_4}$

$(t_1+t_2+t_3+t_4{)}^2-2\sum t_1{t}_2=4\left(\frac{\sum t_1{t}_2{t}_3}{t_1{t}_2{t}_3{t}_4}\right)$

$\Rightarrow \frac{h^2}{4}-2\times 0=4\times \left(\frac{\frac{-k}{4}}{-1}\right)$

$\Rightarrow \frac{h^2}{4}=k$

$\Rightarrow h^2=16k$

so the locus of $(h,k)$ is $x^2=16y$, which is a parabola