Question

The normal to the rectangular hyperbola $xy=c^2$ at the point 't' meets the curve again at a point 't' such that

• A. $t^3{t}^{\prime }=-1$
• B. $t^2{t}^{\prime }=-1$
• C. $t{t}^{\prime }=-1$
• D. None of these

Answer 1

The correct option is A $t^3{t}^{\prime }=-1$

Given: Hyperbola $xy=c^2$--------1

Let there be a point $P(ct,\frac{c}{t})$ from which normal is drawn(in parametric co-ordinates) which is denoted at t.

So, we know that Equation of normal at$(x_1,y_1)$ is

$x{x}_1-y{y}_1=x_1^2-y_1^2$

So, normal at $(ct,\frac{c}{t})$ will be $ctx-\frac{c}{t}y=c^2{t}^2-\frac{c^2}{t^2}$

$\Rightarrow x{t}^3-yt-c{t}^4+c=0$-------------2

The normal again meets the curve at t' i.e. $(c{t}^{{}^{\prime }},\frac{c}{t^{{}^{\prime }}})$

$\Rightarrow It lies on both the normal & the hyperbola (and must satisfy these equations)

Putting $(c{t}^{{}^{\prime }},\frac{c}{t^{{}^{\prime }}})$ in Equation 2

$\Rightarrow c{t}^{{}^{\prime }}{t}^3-\frac{c}{t^{{}^{\prime }}}t-c{t}^4+c=0$

$\Rightarrow c(t^{{}^{\prime }2}{t}^3-t-t^4{t}^{{}^{\prime }}+t^{{}^{\prime }})=0$

$\Rightarrow \left[\right(t^{{}^{\prime }2}{t}^3+t^{{}^{\prime }})+(-t-t^4{t}^{{}^{\prime }}\left)\right]=0$

$\Rightarrow t^{{}^{\prime }}(t^{{}^{\prime }}{t}^3+1)-t(1+t^3{t}^{{}^{\prime }})=0$

$\Rightarrow (t^{{}^{\prime }}-t)=0$ or $(t^{{}^{\prime }}{t}^3+1)=0$

We know $t^{{}^{\prime }}\ne t$ (because if $t^{{}^{\prime }}=t\Rightarrow t$ lies on $t^{{}^{\prime }})$

So, $t^{{}^{\prime }}{t}^3+1=0\Rightarrow t^{{}^{\prime }}{t}^3+1=0$

$\Rightarrow t^{{}^{\prime }}{t}^3=-1$