The line y=x+1 meets the ellipse frac{x^{2}}{ | vedclass

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

Question

(A) The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$,which simplifies to $x^{2}+2y^{2}=4$. Substituting $y=x+1$ into the ellipse equ

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$,which simplifies to $x^{2}+2y^{2}=4$. Substituting $y=x+1$ into the ellipse equation: $x^{2}+2(x+1)^{2}=4$ $x^{2}+2(x^{2}+2x+1)=4$ $3x^{2}+4x-2=0$. Let the roots be $x_{1}$ and $x_{2}$. Then $|x_{1}-x_{2}| = \frac{\sqrt{D}}{|a|} = \frac{\sqrt{16 - 4(3)(-2)}}{3} = \frac{\sqrt{40}}{3}$. The length of the chord $PQ$ is given by $PQ = |x_{1}-x_{2}| \sqrt{1+m^{2}}$,where $m$ is the slope of the line $y=x+1$,so $m=1$. $PQ = \frac{\sqrt{40}}{3} \sqrt{1+1^{2}} = \frac{\sqrt{40}}{3} \sqrt{2} = \frac{\sqrt{80}}{3}$. Since $PQ$ is the diameter of the circle,$2r = PQ = \frac{\sqrt{80}}{3}$,so $r = \frac{\sqrt{80}}{6}$. Therefore,$(3r)^{2} = 9r^{2} = 9 	imes \frac{80}{36} = \frac{80}{4} = 20$.

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter,then $(3r)^{2}$ is equal to

Similar Questions

$A$ focus of an ellipse is at the origin. The directrix is the line $x = 4$ and the eccentricity is $\frac{1}{2}$. Then the length of the semi-major axis is

Find the coordinates of the foci,the vertices,the lengths of the major and minor axes,and the eccentricity of the ellipse $9x^{2} + 4y^{2} = 36$.

The eccentricity of the ellipse $4x^2 + 9y^2 = 36$ is

$A$ tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersects the coordinate axes at $A$ and $B$. Then the locus of the circumcentre of triangle $AOB$ (where $O$ is the origin) is:

The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the straight line $y = mx + c$ intersect in real points only if

Vedclass Test Series

Exam Paper Generator

Online Exam Module