Question

The length of the subnormal of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at P$(x_1,y_1)$ is

• A. $\frac{b^2{x}_1}{a^2}$
• B. $(e^2-1)x_1$
• C. $x_1-\frac{a^2}{x_1}.$
• D. $\frac{b^2{x}_1^2}{a^3}$

Answer 1

Answer: (A), (B)

$\frac{b^2{x}_1}{a^2}$
$(e^2-1)x_1$

For any standard Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Let the Normal at point $P(x_1,y_1)$ meet the $x$-axis at G$(x,0)$,

$C(0,0)$ is the center and the perpendicular drawn from point $P$ to $x-$axis meets the $x-$ axis at $N$

The coordinates of point $N$ are $(x_1,0)$

The equation of tangent at $P(x_1,y_1)$ is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2{e}^2$.....$\left(1\right)$

Point $G$ lies on $x-$axis so its $y$ ordinate is $0$

For finding $x-$ ordinate, putting value of $y=0$ in eq. $\left(1\right)$ we get,

$\Rightarrow$ $x=e^2{x}_1$

So point $G$ is $(e^2{x}_1,0)$

From the figure, we can know that the length of sub-normal for a standard hyperbola is $NG$

Also from figure $NG=CG-CN$

As $C$ is $(0,0)$, $N$ is $(x_1,0)$ and $G$ is $(e^2{x}_1,0)$

$\because$ All the points lie on $x$-axis, in a line.

$\therefore$ $CG=$ $e^2{x}_1$

$CN=x_1$

$\Rightarrow$ $NG=(e^2-1)x_1$,

$\Rightarrow$ for a standard hyperbola we know that $e^2=1+\frac{b^2}{a^2}$ or $e^2-1=\frac{b^2}{a^2}$

Hence the length of sub-normal of a standard hyperbola $NG=\frac{b^2}{a^2}{x}_1$

So length of subnormal $NG=(e^2-1)x_1=\frac{b^2}{a^2}{x}_1$

Hence option $A$ and $B$ both are correct.