Question

If a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes at $M\&N$ and the lines $MP\&NP$ are drawn perpendicular to the axes meeting at $P$, then locus of $P$ is

• A. $(a^2{x}^2-b^2{y}^2)=(a^2-b^2)$
• B. $a^2{x}^2-b^2{y}^2=a^2+b^2$
• C. $a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$
• D. $a^2{x}^2-b^2{y}^2=(a^2-b^2{)}^2$

Answer 1

The correct option is C $a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$

The equation of any normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{is}ax\cos \varphi +\text{by cot}\varphi =a^2+b^2\cdots \left(1\right)$

Let the normal meets $x$ -axis at $M$ & $y$ - axis at $N$ respectivley, So
$M=(\left(\frac{a^2+b^2}{a}\right)\sec \varphi ,0)\&N=(0,\left(\frac{a^2+b^2}{b}\right)\tan \varphi )$

Let locus of $P$ be $(\alpha ,\beta )$
Since $PM\&PN$ are perpendicular to the axes, co - ordinates of $P$ is
$(\left(\frac{a^2+b^2}{a}\right)\sec \varphi ,\left(\frac{a^2+b^2}{b}\right)\tan \varphi )$
$\therefore \alpha =\left(\frac{a^2+b^2}{a}\right)\sec \varphi \&\beta =\left(\frac{a^2+b^2}{b}\right)\tan \varphi$
$\Rightarrow \alpha \left(\frac{a}{a^2+b^2}\right)=\sec \varphi \&\beta \left(\frac{b}{a^2+b^2}\right)=\tan \varphi$

As ${\sec }^2\varphi -{\tan }^2\varphi =1$
${\alpha }^2{\left(\frac{a}{a^2+b^2}\right)}^2-{\beta }^2{\left(\frac{b}{a^2+b^2}\right)}^2=1$
$\Rightarrow {\alpha }^2{a}^2-{\beta }^2{b}^2=(a^2+b^2{)}^2$
Hence locus of $(\alpha ,\beta )$ is
$a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$