Question

Let P (6, 3) be the point on the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. If the normal at the point P intersects at the X-axis at (9, 0), then the eccentricity of the hyperbola is
(a) $\sqrt{\dfrac{5}{2}}$
(b) $\sqrt{\dfrac{3}{2}}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$

Answer 1

Hint: We use the normal equation of hyperbola which is given by ${{a}^{2}}{{y}_{1}}(x-{{x}_{1}})+{{b}^{2}}{{x}_{1}}(y-{{y}_{1}})=0$ where ${{x}_{1}}$ and ${{y}_{1}}$ are the coordinates of point P (6, 3).
Complete step-by-step answer:
The equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is the equation of family of hyperbolas. By considering a=b=1 (in red colour) and substituting a=b=2 (in black colour) in the equation we get the following diagram.

As we can see that the equation of hyperbola is given and since the point P(6,3) lies on this equation so we can substitute the value P(6,3) in equation of hyperbola as such in place of x and y in $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
Thus we get,
$\begin{align}
  & \dfrac{{{\left( 6 \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( 3 \right)}^{2}}}{{{b}^{2}}}=1 \\
 & \dfrac{36}{{{a}^{2}}}-\dfrac{9}{{{b}^{2}}}=1 \\
\end{align}$
Now we use the normal equation of hyperbola which is ${{a}^{2}}{{y}_{1}}(x-{{x}_{1}})+{{b}^{2}}{{x}_{1}}(y-{{y}_{1}})=0$ and substitute the value of P to it. This implies $3{{a}^{2}}(x-{{x}_{1}})+6{{b}^{2}}(y-{{y}_{1}})=0....(i)$
This implies that $3{{a}^{2}}(x-{{x}_{1}})+6{{b}^{2}}(y-{{y}_{1}})=0$
Proceeding further we get,
$\begin{align}
  & 3{{a}^{2}}(x-{{x}_{1}})+6{{b}^{2}}(y-{{y}_{1}})=0 \\
 & 3{{a}^{2}}(x-6)+6{{b}^{2}}(y-3)=0 \\
 & (3{{a}^{2}}x-18{{a}^{2}})+(6{{b}^{2}}y-18{{b}^{2}})=0 \\
 & (3{{a}^{2}}x-18{{a}^{2}})=-(6{{b}^{2}}y-18{{b}^{2}}) \\
\end{align}$
This further implies $3{{a}^{2}}x-18{{a}^{2}}=-6{{b}^{2}}y+18{{b}^{2}}$
As it is clear from the question that the normal at the point P intersects at the X-axis at (9, 0), so we can substitute the point (9, 0) in equation $(i)$ in place of $\left( x,y \right)$ . Thus we have,
$\begin{align}
  & 3{{a}^{2}}\left( 9 \right)-18{{a}^{2}}=-6{{b}^{2}}\left( 0 \right)+18{{b}^{2}} \\
 & 3{{a}^{2}}\left( 9 \right)-18{{a}^{2}}=18{{b}^{2}} \\
 & 27{{a}^{2}}-18{{a}^{2}}=18{{b}^{2}} \\
 & 9{{a}^{2}}=18{{b}^{2}} \\
 & {{a}^{2}}=2{{b}^{2}} \\
\end{align}$
This gives the simplest form of the equation and that is $a=\sqrt{2}b....(ii)$
 Now we use the formula of eccentricity of hyperbola which is $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}....(iii)$ where e is called eccentricity of hyperbola.
Now we substitute equation $(ii)$ in $(iii)$ and we get,
$\begin{align}
  & \dfrac{\sqrt{{{\left( \sqrt{2}b \right)}^{2}}+{{b}^{2}}}}{\sqrt{2}b}=\dfrac{\sqrt{2{{b}^{2}}+{{b}^{2}}}}{\sqrt{2}b} \\
 & \dfrac{\sqrt{{{\left( \sqrt{2}b \right)}^{2}}+{{b}^{2}}}}{\sqrt{2}b}=\dfrac{\sqrt{3{{b}^{2}}}}{\sqrt{2}b} \\
 & \dfrac{\sqrt{{{\left( \sqrt{2}b \right)}^{2}}+{{b}^{2}}}}{\sqrt{2}b}=\dfrac{\sqrt{3}b}{\sqrt{2}b} \\
\end{align}$
Now we cancel b from numerator and denominator. This implies $\dfrac{\sqrt{3}b}{\sqrt{2}b}=\sqrt{\dfrac{3}{2}}$
Hence the correct option is (b).

Note: By substituting the point only, we can get the simplest form of the equation instead of solving without substituting. The substitution in the right equation will lead to the desired result. Squaring a negative sign will imply a positive sign. Moreover, try to cancel common terms in numerator and denominator to get a simple form. This results in reaching the right answer.