What will be the equation of that chord of el | vedclass

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

Question

(d) Let required chord meets to ellipse on the points $P$ and $Q$, whose co-ordinates are $({x_1},{y_1})$ and $({x_2},{y_2})$ respectively.

$	here

Vedclass · Accepted Answer

$x + 2y + 4$

Vedclass · Answer

(d) Let required chord meets to ellipse on the points $P$ and $Q$, whose co-ordinates are $({x_1},{y_1})$ and $({x_2},{y_2})$ respectively.

$	herefore $point $(2, 1)$ is mid point of chord $PQ$

$	herefore $ $2 = \frac{1}{2}$ $({x_1} + {x_2})$ or ${x_1} + {x_2} = 4$

and $1 = \frac{1}{2}({y_1} + {y_2})\,$ or ${y_1} + {y_2} = 2$

Again point $({x_1},{y_1})$ and $({x_2},{y_2})$ are situated on ellipse

$	herefore $ $\frac{{x_1^2}}{{36}} + \frac{{y_1^2}}{9} = 1$ and $\frac{{x_2^2}}{{36}} + \frac{{y_2^2}}{9} = 1$

On subtracting, $\frac{{x_2^2 - x_1^2}}{{36}} + \frac{{y_2^2 - y_1^2}}{9} = 0$

or $\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = $$\frac{{ - ({x_2} + {x_1})}}{{4({y_2} + {y_1})}} = \frac{{ - 4}}{{4 	imes 2}} = - \frac{1}{2}$

$	herefore $ Gradient of chord $PQ$ $ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = - \frac{1}{2}$

Therefore, required equation of chord $PQ$ is as follows $y-1 = - \frac{1}{2}(x - 2)$ or $x + 2y = 4.$

What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

Similar Questions

The distance of the point $'\theta '$on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ from a focus is

Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is

The ellipse ${x^2} + 4{y^2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in trun is inscribed in another ellipse that passes through the point $(4,0) $ . Then the equation of the ellipse is :

The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to:

Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is