The latus rectum of the conic satisfying the | vedclass

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

Question

(C) The given differential equation is $x dy + y dx = 0$,which can be written as $d(xy) = 0$.Integrating both sides,we get $xy = c$.Since the curve pa

Vedclass · Accepted Answer

$8 \sqrt{2}$

Vedclass · Answer

(C) The given differential equation is $x dy + y dx = 0$,which can be written as $d(xy) = 0$.Integrating both sides,we get $xy = c$.Since the curve passes through $(2, 8)$,we have $2 	imes 8 = c$,so $c = 16$.The equation of the conic is $xy = 16$,which represents a rectangular hyperbola.For a rectangular hyperbola $xy = c$,the equation can be transformed to the standard form $X^2 - Y^2 = a^2$ by rotating the axes by $45^\circ$.Here,$xy = 16$ is equivalent to $\frac{x^2}{32} - \frac{y^2}{32} = -1$ (or $Y^2 - X^2 = 32$ in rotated coordinates).Comparing with $Y^2 - X^2 = a^2$,we have $a^2 = 32$,so $a = 4\sqrt{2}$.The length of the latus rectum for a rectangular hyperbola $xy = c$ is $2\sqrt{2} 	imes 	ext{semi-transverse axis} 	imes \sqrt{2} = 4a$ is not correct; rather,for $xy = c$,the latus rectum is $2\sqrt{2} 	imes \sqrt{2c} = 4\sqrt{c}$.Alternatively,for $xy = 16$,$a = 4\sqrt{2}$. The latus rectum is $2a = 2(4\sqrt{2}) = 8\sqrt{2}$.

The latus rectum of the conic satisfying the differential equation $x dy + y dx = 0$ and passing through the point $(2, 8)$ is:

Similar Questions

If $y = e^{-mx}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + 3y = 0$, then the values of $m$ are

If $y = At^2 + \frac{B}{t}$ ($A, B$ are parameters) is the general solution of the differential equation $f(t) y''(t) + g(t) y'(t) + h(t) y = 0$,then $2 f(t) + t^2 h(t) =$

If the slope of the tangent on a curve at any point $(x, y)$ is equal to $\frac{y^2-x^2}{2xy}$,then the equation of the normal at the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module