The average length of all vertical chords of | vedclass

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

Question

(A) The length of a vertical chord at a given $x$ is $L(x) = 2y = 2\frac{b}{a}\sqrt{x^2-a^2}$.The average length $A_L$ is given by $\frac{1}{2a-a} \in

Vedclass · Accepted Answer

$b\{2 \sqrt{3}-\ln(2+\sqrt{3})\}$

Vedclass · Answer

(A) The length of a vertical chord at a given $x$ is $L(x) = 2y = 2\frac{b}{a}\sqrt{x^2-a^2}$.The average length $A_L$ is given by $\frac{1}{2a-a} \int_a^{2a} L(x) dx = \frac{1}{a} \int_a^{2a} 2\frac{b}{a}\sqrt{x^2-a^2} dx$.$A_L = \frac{2b}{a^2} \int_a^{2a} \sqrt{x^2-a^2} dx$.Using the integral formula $\int \sqrt{x^2-a^2} dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|$, we evaluate from $a$ to $2a$:$A_L = \frac{2b}{a^2} [\frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln(x+\sqrt{x^2-a^2})]_a^{2a}$.At $x=2a$: $\frac{2a}{2}\sqrt{4a^2-a^2} - \frac{a^2}{2}\ln(2a+\sqrt{3}a) = a^2\sqrt{3} - \frac{a^2}{2}\ln(a(2+\sqrt{3}))$.At $x=a$: $\frac{a}{2}\sqrt{0} - \frac{a^2}{2}\ln(a) = - \frac{a^2}{2}\ln(a)$.Subtracting these: $A_L = \frac{2b}{a^2} [a^2\sqrt{3} - \frac{a^2}{2}\ln(a(2+\sqrt{3})) + \frac{a^2}{2}\ln(a)] = \frac{2b}{a^2} [a^2\sqrt{3} - \frac{a^2}{2}\ln(2+\sqrt{3})] = b[2\sqrt{3}-\ln(2+\sqrt{3})]$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

The average length of all vertical chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ for $a \leq x \leq 2a$ is:

Similar Questions

$\int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx =$

$\int_0^a x^2 (a^2 - x^2)^{3/2} dx = $

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:
$t \ (\text{in second})$ $0$ $2$ $4$ $6$ $8$ $10$
$v \ (\text{in m/s})$ $0$ $12$ $16$ $20$ $35$ $60$

If $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$,then $(b-1)^2 =$

The approximate value of $\int_2^{10} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module