Let P(6, 3) be a point on the hyperbola frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} + \frac{b^

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.Substituting $(x_1, y_1) = (6, 3)$,we get $\frac{a^2x}{6} + \frac{b^2y}{3} = a^2 + b^2$.Since the normal passes through $(9, 0)$,we substitute $x = 9$ and $y = 0$:$\frac{a^2(9)}{6} + 0 = a^2 + b^2$$\frac{3a^2}{2} = a^2 + b^2$$\frac{1}{2}a^2 = b^2 \Rightarrow a^2 = 2b^2$.The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.Substituting $a^2 = 2b^2$:$e = \sqrt{1 + \frac{b^2}{2b^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$.

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

Similar Questions

The value of the series $\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$ is

$A$ proton, when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of:

$Assertion (A)$: van der Waals' forces are responsible for chemisorption. $Reason (R)$: High temperature is favourable for chemisorption. The correct answer is

Which of the following organelles contains hydrolytic enzymes?

The maximum area of a right-angled triangle with hypotenuse $h$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module