The angle between the pair of tangents drawn | vedclass

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

Question

(C) The equation of the pair of tangents from point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $SS_1 = T^2$.For t

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{12}{\sqrt{5}}ight)$

Vedclass · Answer

(C) The equation of the pair of tangents from point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $SS_1 = T^2$.For the ellipse $3x^2 + 2y^2 - 5 = 0$ and point $(1, 2)$:$S = 3x^2 + 2y^2 - 5$$S_1 = 3(1)^2 + 2(2)^2 - 5 = 3 + 8 - 5 = 6$$T = 3x(1) + 2y(2) - 5 = 3x + 4y - 5$Substituting into $SS_1 = T^2$:$6(3x^2 + 2y^2 - 5) = (3x + 4y - 5)^2$$18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$$9x^2 - 24xy - 4y^2 + 30x + 40y - 55 = 0$This is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a = 9, h = -12, b = -4$.The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.$	an 	heta = \left| \frac{2\sqrt{(-12)^2 - (9)(-4)}}{9 - 4} ight| = \left| \frac{2\sqrt{144 + 36}}{5} ight| = \frac{2\sqrt{180}}{5} = \frac{2 	imes 6\sqrt{5}}{5} = \frac{12\sqrt{5}}{5} = \frac{12}{\sqrt{5}}$.Therefore,$	heta = 	an^{-1}\left(\frac{12}{\sqrt{5}}ight)$.

The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3x^2 + 2y^2 = 5$ is:

Similar Questions

The focus of an ellipse is at the origin. The line $x = 4$ is a directrix and the eccentricity is $1/2$. Find the length of the major axis. (in $/3$)

Number of points on the ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$ from which a pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

An ellipse has its major axis along the $y$-axis and its minor axis along the $x$-axis. If the length of its latus rectum is $\frac{2}{3}$ times the length of its minor axis,then the eccentricity of the ellipse is:

The eccentricity of the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$ is

Let the foci of the ellipse $\frac{x^{2}}{9}+y^{2}=1$ subtend a right angle at a point $P$. Then, the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module