The locus of the midpoints of the parallel ch | vedclass

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

Question

(D) The equation of a chord of the hyperbola $xy = c^2$ with midpoint $(h, k)$ is given by $T = S_1$,which is $xh + yk = 2c^2$ is incorrect; the corre

Vedclass · Accepted Answer

$my + x = 0$

Vedclass · Answer

(D) The equation of a chord of the hyperbola $xy = c^2$ with midpoint $(h, k)$ is given by $T = S_1$,which is $xh + yk = 2c^2$ is incorrect; the correct formula for the chord with midpoint $(h, k)$ for $xy = c^2$ is $xh + yk = 2h k$.Comparing this to the slope-intercept form $y = mx + c'$,we have $y = -(\frac{h}{k})x + 2h$.Given the gradient of the chord is $m$,we have $m = -\frac{h}{k}$.Thus,$h = -mk$.Replacing $(h, k)$ with the general point $(x, y)$,we get $x = -my$,which implies $x + my = 0$.

The locus of the midpoints of the parallel chords with gradient $m$ of the rectangular hyperbola $xy = c^2$ is

Similar Questions

If the angle between the asymptotes of a hyperbola is $30^{\circ}$,then its eccentricity is

If the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \frac{5}{4}$ and the length of the latus rectum equal to $9$,then $ab$ is equal to

$A$ square $ABCD$ has all its vertices on the curve $x^{2}y^{2}=1$. The midpoints of its sides also lie on the same curve. Then,the square of the area of $ABCD$ is

Area of the triangle formed by the lines $x - y = 0$,$x + y = 0$ and any tangent to the hyperbola $x^2 - y^2 = a^2$ is:

$A$ hyperbola has its centre at the origin,passes through the point $(4, 2)$ and has a transverse axis of length $4$ along the $x$-axis. Then the eccentricity of the hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module