Question

Let P(6,3) be a point on the hyperbola \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\] the normal at the point P intersects the X-axis at (9, 0), then the eccentricity of the hyperbola is______.

विकल्प

• \[\sqrt{\frac{5}{2}}\]

• \[\sqrt{\frac{3}{2}}\]

• \[\sqrt{2}\]

• \[\sqrt{3}\]

Answer 1

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

Explanation:

Equation of normal at \[(x_{1},y_{1})\] to hyperbola

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\mathrm{is}\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2\]

Given, \[(x_1,y_1)=(6,3)\]

\[\therefore\] Equation is \[\frac{a^{2}x}{6}+\frac{b^{2}y}{3}=a^{2}+b^{2}.\]

It intersection X-axis at (9, 0).

\[\therefore\quad\frac{a^2(9)}{6}+\frac{b^2(0)}{3}=a^2+b^2\]

\[\Rightarrow\quad3a^2=2a^2+2b^2\]

\[\Rightarrow\quad a^2=2b^2\Longrightarrow\frac{a^2}{b^2}=2\]

For Hyperbola, \[b^2=a^2(e^2-1)\]

                         \[\frac{1}{2}=e^2-1\]

                         \[e^2=1+\frac{1}{2}\]

                         \[e=\sqrt{\frac{3}{2}}\]