The normal drawn at the point left(sqrt{9} co | vedclass

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

Question

(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$. The point given is $(x_1, y_1) = \left(3 \cos \f

Vedclass · Accepted Answer

$\left(\sqrt{\frac{2}{9}}, 0\right)$

Vedclass · Answer

(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$. The point given is $(x_1, y_1) = \left(3 \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right) = \left(\frac{3}{\sqrt{2}}, \sqrt{\frac{7}{2}}\right)$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Substituting the values: $\frac{9x}{3/\sqrt{2}} - \frac{7y}{\sqrt{7/2}} = 9 - 7$. This simplifies to $3\sqrt{2}x - \sqrt{14}y = 2$. To find the intersection with the major axis ($x$-axis),set $y = 0$: $3\sqrt{2}x = 2 \implies x = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$. Thus,the point of intersection is $\left(\sqrt{\frac{2}{9}}, 0\right)$.

The normal drawn at the point $\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{7} = 1$ intersects its major axis at the point

Similar Questions

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 locus of the midpoint of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2y^{2}=4$ is an ellipse with eccentricity:

The curve represented by $x = 5(\cos t + \sin t)$ and $y = 3(\cos t - \sin t)$ is (where $t$ is a parameter):

The eccentricity of an ellipse is $2/3$,the length of the latus rectum is $5$,and the centre is $(0, 0)$. The equation of the ellipse is:

$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3x^2 + 4y^2 = 12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$,then $a =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module