A point on the ellipse 4x^2 + 9y^2 = 36,where | vedclass

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

Question

(A) The given ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The normal is p

Vedclass · Accepted Answer

$\left( \frac{9}{5}, \frac{8}{5} \right)$

Vedclass · Answer

(A) The given ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The normal is parallel to the line $4x - 2y - 5 = 0$,so the slope of the normal is $m_n = \frac{-4}{-2} = 2$. The slope of the tangent at that point is $m_t = -\frac{1}{m_n} = -\frac{1}{2}$. The equation of the tangent with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$. The point of contact is $\left( -\frac{a^2m}{c}, \frac{b^2}{c} \right)$ where $c = \pm \sqrt{a^2m^2 + b^2}$. Here $m = -\frac{1}{2}$,so $c = \sqrt{9(-\frac{1}{2})^2 + 4} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$. The point of contact is $\left( -\frac{9(-1/2)}{5/2}, \frac{4}{5/2} \right) = \left( \frac{9/2}{5/2}, \frac{8}{5} \right) = \left( \frac{9}{5}, \frac{8}{5} \right)$. Since the normal is parallel to $4x - 2y - 5 = 0$,the point is $\left( \frac{9}{5}, \frac{8}{5} \right)$.

$A$ point on the ellipse $4x^2 + 9y^2 = 36$,where the normal is parallel to the line $4x - 2y - 5 = 0$,is

Similar Questions

If the maximum distance of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$,where $b < 2$,from the origin is $1$,then the eccentricity of the ellipse is:

The eccentricity of the ellipse with minor axis $2b$,if the line segment joining the foci subtends an angle $2\alpha$ at the upper vertex,is equal to

The angle between the tangents drawn from a point $(-3, 2)$ to the ellipse $4x^2 + 9y^2 - 36 = 0$ is

Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.
Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.
Which of the above statement$(s)$ is (are) true?

If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at

Vedclass Test Series

Exam Paper Generator

Online Exam Module