The midpoint of the chord 4x - 3y = 5 of the | vedclass

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

Question

(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and

Vedclass · Accepted Answer

$(2, 1)$

Vedclass · Answer

(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and $S_1 = 2h^2 - 3k^2 - 12$. Thus,the equation of the chord is $2xh - 3yk = 2h^2 - 3k^2$. Comparing this with the given chord equation $4x - 3y = 5$,we get: $\frac{2h}{4} = \frac{-3k}{-3} = \frac{2h^2 - 3k^2}{5}$. From $\frac{2h}{4} = k$,we have $k = \frac{h}{2}$. Substituting $k = \frac{h}{2}$ into $\frac{2h}{4} = \frac{2h^2 - 3(h/2)^2}{5}$: $\frac{h}{2} = \frac{2h^2 - \frac{3h^2}{4}}{5} = \frac{5h^2}{20} = \frac{h^2}{4}$. So,$\frac{h}{2} = \frac{h^2}{4}$ $\Rightarrow 2h = h^2$ $\Rightarrow h(h - 2) = 0$. Since the chord exists,$h 
eq 0$ (as $h=0$ gives $k=0$,which does not satisfy the chord equation $4(0)-3(0)=5$). Thus,$h = 2$. Then $k = \frac{2}{2} = 1$. The midpoint is $(2, 1)$.

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

Similar Questions

Find the slope of the tangent to the hyperbola $2x^2 - 3y^2 = 6$ at the point $(3, 2)$.

$A$ mirror in the first quadrant is in the shape of a hyperbola whose equation is $xy=1$. $A$ light source in the second quadrant emits a beam of light that hits the mirror at the point $(2, 1/2)$. If the reflected ray is parallel to the $Y$-axis,the slope of the incident beam is

The locus of the point of intersection of the tangents drawn at the extremities of a normal chord of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

The equation of a line passing through the centre of a rectangular hyperbola is $x - y - 1 = 0$. If one of the asymptotes is $3x - 4y - 6 = 0$,the equation of the other asymptote is:

The directrix of the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module