Question

Paragraph :
Let $S$ and $S^{\prime }$ be the foci on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $P(x_1,y_1)$ be an arbitrary point as shown.
Hence, find the value of $\tan \varphi$.

• A. $\frac{a^2}{ae{y}_1}$
• B. $\frac{b^2}{ae{y}_1}$
• C. $\frac{b^2}{ae{x}_1}$
• D. $\frac{a^2}{ae{x}_1}$

Answer 1

The correct option is B $\frac{b^2}{ae{y}_1}$

$\tan \varphi$ is the acute angle between the lines $PS$ and the tangent at $P$.
i.e. $\tan \varphi =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$=\left|\frac{\frac{y_1}{x_1-ae}-\frac{b^2{x}_1}{a^2{y}_1}}{1+\left(\frac{y_1}{x_1-ae}\right)\left(\frac{b^2{x}_1}{a^2{y}_1}\right)}\right|$
$=\left|\frac{a^2{y}_1^2-b^2{x}_1(x_1-ae)}{(x_1-ae)a^2{y}_1+b^2{x}_1{y}_1}\right|$
$=\left|\frac{a^2{y}_1^2-b^2{x}_1^2+a{b}^2{x}_{1}e}{(a^2+b^2)x_1{y}_1-a^{3}e{y}_1}\right|$
But $a^2{y}_1^2-b^2{x}_1^2=-a^2{b}^2$ and $a^2+b^2=a^2{e}^2$,

Substituting these gives :
$\tan \varphi =\left|\frac{-a^2{b}^2+a{b}^2{x}_{1}e}{a^2{e}^2{x}_1{y}_1-a^{3}e{y}_1}\right|=\left|\frac{b^2(a{x}_{1}e-a^2)}{ae{y}_1(a{x}_{1}e-a^2)}\right|=\frac{b^2}{ae{y}_1}$.
Hence, option B is correct.