If the line x - 2y = 12 is tangent to the ell | vedclass

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

Question

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} =

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Given the point $(3, -4.5)$,the tangent equation is $\frac{3x}{a^2} - \frac{4.5y}{b^2} = 1$.Comparing this with the given tangent line $x - 2y = 12$,we rewrite the given line as $\frac{x}{12} - \frac{2y}{12} = 1$,which simplifies to $\frac{x}{12} - \frac{y}{6} = 1$.Equating the coefficients: $\frac{3}{a^2} = \frac{1}{12}$ $\Rightarrow a^2 = 36$ $\Rightarrow a = 6$.And $\frac{4.5}{b^2} = \frac{1}{6} \Rightarrow b^2 = 4.5 	imes 6 = 27$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 27}{6} = \frac{54}{6} = 9$.

If the line $x - 2y = 12$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(3, -4.5)$,then the length of the latus rectum of the ellipse is:

Similar Questions

Find the equation of the ellipse $(a > b)$ whose distance between the foci is $8$ and the distance between the directrices is $18$.

The equation of the ellipse whose one of the vertices is $(0, 7)$ and the corresponding directrix is $y = 12$ is:

The equation of the ellipse having a vertex at $(6,1)$,a focus at $(4,1)$ and the eccentricity $e = \frac{3}{5}$ is

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

If $\theta$ is the eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ such that the distance of the point from the center is $5$, then $\theta = ......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module