The equation of the line passing through the | vedclass

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

Question

(A) The slope $m$ of the line passing through $(ct_1, c/t_1)$ and $(ct_2, c/t_2)$ is given by: $m = \frac{\frac{c}{t_2} - \frac{c}{t_1}}{ct_2 - ct_1}

Vedclass · Accepted Answer

$x+t_1 t_2 y=c\left(t_1+t_2\right)$

Vedclass · Answer

(A) The slope $m$ of the line passing through $(ct_1, c/t_1)$ and $(ct_2, c/t_2)$ is given by: $m = \frac{\frac{c}{t_2} - \frac{c}{t_1}}{ct_2 - ct_1} = \frac{c(t_1 - t_2)}{t_1 t_2} \cdot \frac{1}{c(t_2 - t_1)} = -\frac{1}{t_1 t_2}$ Using the point-slope form $(y - y_1) = m(x - x_1)$ with point $(ct_1, c/t_1)$: $y - \frac{c}{t_1} = -\frac{1}{t_1 t_2}(x - ct_1)$ Multiplying both sides by $t_1 t_2$: $t_1 t_2 y - ct_2 = -(x - ct_1)$ $t_1 t_2 y - ct_2 = -x + ct_1$ $x + t_1 t_2 y = c(t_1 + t_2)$

The equation of the line passing through the points $\left(ct_1, \frac{c}{t_1}\right)$ and $\left(ct_2, \frac{c}{t_2}\right)$ is

Similar Questions

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

The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2x^2-y^2=4$ is

If the latus rectum of a hyperbola subtends an angle of $120^{\circ}$ at its centre,then its eccentricity is

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module