The focal distances of the point left(frac{4} | vedclass

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

Question

(D) Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $b^2 > a^2$ $(9 > 4)$,this is a vertical ellipse with $a^2 = 4$ and $b^2 = 9

Vedclass · Accepted Answer

$4, 2$

Vedclass · Answer

(D) Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $b^2 > a^2$ $(9 > 4)$,this is a vertical ellipse with $a^2 = 4$ and $b^2 = 9$. The eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$,so $4 = 9(1 - e^2)$ $\Rightarrow 1 - e^2 = \frac{4}{9}$ $\Rightarrow e^2 = \frac{5}{9}$ $\Rightarrow e = \frac{\sqrt{5}}{3}$. The foci are $(0, \pm be) = (0, \pm 3 	imes \frac{\sqrt{5}}{3}) = (0, \pm \sqrt{5})$. Let $S = (0, \sqrt{5})$ and $S' = (0, -\sqrt{5})$. Let $P = \left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}ight)$. The focal distances are $PS$ and $PS'$. $PS = \sqrt{\left(\frac{4}{\sqrt{5}} - 0ight)^2 + \left(\frac{3}{\sqrt{5}} - \sqrt{5}ight)^2} = \sqrt{\frac{16}{5} + \left(\frac{3-5}{\sqrt{5}}ight)^2} = \sqrt{\frac{16}{5} + \frac{4}{5}} = \sqrt{\frac{20}{5}} = 2$. $PS' = \sqrt{\left(\frac{4}{\sqrt{5}} - 0ight)^2 + \left(\frac{3}{\sqrt{5}} + \sqrt{5}ight)^2} = \sqrt{\frac{16}{5} + \left(\frac{3+5}{\sqrt{5}}ight)^2} = \sqrt{\frac{16}{5} + \frac{64}{5}} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4$. Thus,the focal distances are $4$ and $2$.

The focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ are

Similar Questions

If $x+\sqrt{3} y=3$ is the tangent to the ellipse $2 x^2+3 y^2=k$ at a point $P$,then the equation of the normal to this ellipse at $P$ is

Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$. Then the maximum perpendicular distance of $P$ from the line $x - y = 10$ is (in $\sqrt{2}$)

If tangents are drawn from point $P(3 \sin \theta + 4 \cos \theta, 3 \cos \theta - 4 \sin \theta)$ where $\theta = \frac{\pi}{8}$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,then the angle between the tangents is:

The values that $m$ can take,so that the straight line $y=4x+m$ touches the curve $x^2+4y^2=4$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module