The equation of a chord AB of an ellipse 2x^2 | vedclass

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

Question

(B) The equation of the ellipse is $2x^2 + y^2 = 1$. The equation of the chord is $x - y + 1 = 0$,which implies $y = x + 1$.Substituting $y = x + 1$ i

Vedclass · Accepted Answer

$\operatorname{Tan}^{-1}(2)$

Vedclass · Answer

(B) The equation of the ellipse is $2x^2 + y^2 = 1$. The equation of the chord is $x - y + 1 = 0$,which implies $y = x + 1$.Substituting $y = x + 1$ into the ellipse equation: $2x^2 + (x + 1)^2 = 1$.$2x^2 + x^2 + 2x + 1 = 1 \implies 3x^2 + 2x = 0$.$x(3x + 2) = 0$,so $x_1 = 0$ and $x_2 = -\frac{2}{3}$.Corresponding $y$ values: $y_1 = 0 + 1 = 1$ and $y_2 = -\frac{2}{3} + 1 = \frac{1}{3}$.Thus,the points are $A(0, 1)$ and $B(-\frac{2}{3}, \frac{1}{3})$.The slopes of $OA$ and $OB$ are $m_1 = \frac{1-0}{0-0} = \infty$ and $m_2 = \frac{1/3 - 0}{-2/3 - 0} = -\frac{1}{2}$.The angle $	heta$ between lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Since $m_1 = \infty$,$	an 	heta = |\frac{1}{m_2}| = |\frac{1}{-1/2}| = 2$.Therefore,$	heta = \operatorname{Tan}^{-1}(2)$.

The equation of a chord $AB$ of an ellipse $2x^2 + y^2 = 1$ is $x - y + 1 = 0$. If $O$ is the origin,then $\angle AOB =$

Similar Questions

If the point $P$ on the curve $4x^{2} + 5y^{2} = 20$ is farthest from the point $Q(0, -4)$,then $PQ^{2}$ is equal to:

The eccentricity of an ellipse whose latus rectum is equal to the distance between its two foci is:

The eccentricity of the ellipse $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$ is

The focus of an ellipse is at the origin. The line $x = 4$ is a directrix and the eccentricity is $1/2$. Find the length of the major axis. (in $/3$)

Tangents are drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at all the four ends of its latus rectum. The area (in sq units) of the quadrilateral formed by these tangents is

Vedclass Test Series

Exam Paper Generator

Online Exam Module