Find the equation of the tangent to the ellip | vedclass

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

Question

(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$.Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Comparing this with the standa

Vedclass · Accepted Answer

$\frac{x}{3} - \frac{y}{2} = 1$

Vedclass · Answer

(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$.Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$ is given by $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (3, -2)$,$a^2 = 9$,and $b^2 = 4$ into the formula:$\frac{x(3)}{9} + \frac{y(-2)}{4} = 1$$\frac{3x}{9} - \frac{2y}{4} = 1$$\frac{x}{3} - \frac{y}{2} = 1$.

Find the equation of the tangent to the ellipse $4x^2 + 9y^2 = 36$ at the point $(3, -2)$.

Similar Questions

If the eccentricity of an ellipse is $1/\sqrt{2}$,then its latus rectum is equal to its

If $P$ is any point on the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ and $S$ and $S^{\prime}$ are the foci,then $PS + PS^{\prime}$ is equal to

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :

The eccentricity of an ellipse,with its centre at the origin,is $\frac{1}{2}$. If one of the directrices is $x = 4$,then the equation of the ellipse is

An ellipse is drawn by taking a diameter of the circle $(x - 1)^2 + y^2 = 1$ as its semi-minor axis and a diameter of the circle $x^2 + (y - 2)^2 = 4$ as its semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes,then the equation of the ellipse is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module