If the midpoint of a chord of the ellipse fra | vedclass

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

Question

(B) The equation of a chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with midpoint $(x_1, y_1)$ is given by $T=S_1$,where $T = \frac{xx_1}{a

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) The equation of a chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with midpoint $(x_1, y_1)$ is given by $T=S_1$,where $T = \frac{xx_1}{a^2}+\frac{yy_1}{b^2}$ and $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$.Given the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and midpoint $(\sqrt{2}, 4/3)$,the equation of the chord is:$\frac{x(\sqrt{2})}{9}+\frac{y(4/3)}{4} = \frac{(\sqrt{2})^2}{9}+\frac{(4/3)^2}{4}$$\frac{\sqrt{2}x}{9}+\frac{y}{3} = \frac{2}{9}+\frac{16/9}{4} = \frac{2}{9}+\frac{4}{9} = \frac{6}{9} = \frac{2}{3}$Multiplying by $9$,we get $\sqrt{2}x+3y=6$,or $y = \frac{6-\sqrt{2}x}{3}$.Substituting this into the ellipse equation $\frac{x^2}{9}+\frac{y^2}{4}=1$:$4x^2 + 9y^2 = 36$$4x^2 + 9\left(\frac{6-\sqrt{2}x}{3}\right)^2 = 36$$4x^2 + (6-\sqrt{2}x)^2 = 36$$4x^2 + 36 + 2x^2 - 12\sqrt{2}x = 36$$6x^2 - 12\sqrt{2}x = 0$$6x(x-2\sqrt{2}) = 0$,so $x_1=0$ and $x_2=2\sqrt{2}$.Corresponding $y$ values are $y_1 = \frac{6-0}{3} = 2$ and $y_2 = \frac{6-\sqrt{2}(2\sqrt{2})}{3} = \frac{6-4}{3} = 2/3$.The length of the chord is the distance between $(0, 2)$ and $(2\sqrt{2}, 2/3)$:$L = \sqrt{(2\sqrt{2}-0)^2 + (2/3-2)^2} = \sqrt{8 + (-4/3)^2} = \sqrt{8 + 16/9} = \sqrt{\frac{72+16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3}$.Comparing this with $\frac{2\sqrt{\alpha}}{3}$,we get $\alpha = 22$.

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4/3)$,and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$,then $\alpha$ is :

Similar Questions

The point $(1,3)$ with respect to the ellipse $4x^2+9y^2-16x-54y+61=0$ lies

The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is

The equation $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$ represents:

If the normal at one end of a latus rectum of the ellipse $\frac{x^2}{32}+\frac{y^2}{b^2}=1$ passes through one end of the minor axis,then $\frac{e^4}{1-e^2}=$ (Here $e$ is the eccentricity of the ellipse)

The maximum length of a chord of the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,such that the eccentric angles of its extremities differ by $\frac{\pi}{2}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module