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 ellipse is $S: 3x^2 + 2y^2 - 5 = 0$. For a point $(x_1, y_1) = (1, 2)$,the equation of the pair of tangents is given by $SS_1

Vedclass · Accepted Answer

$	an^{-1}(12/\sqrt{5})$

Vedclass · Answer

(C) The equation of the ellipse is $S: 3x^2 + 2y^2 - 5 = 0$. For a point $(x_1, y_1) = (1, 2)$,the equation of the pair of tangents is given by $SS_1 = T^2$. Here,$S_1 = 3(1)^2 + 2(2)^2 - 5 = 3 + 8 - 5 = 6$. The tangent $T$ at $(1, 2)$ is $3(1)x + 2(2)y - 5 = 3x + 4y - 5$. Thus,$(3x^2 + 2y^2 - 5)(6) = (3x + 4y - 5)^2$. Expanding this: $18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$. Rearranging into the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$: $9x^2 - 24xy - 4y^2 + 30x + 40y - 55 = 0$. Here,$a = 9$,$h = -12$,and $b = -4$. The angle $	heta$ between the 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(6\sqrt{5})}{5} = \frac{12\sqrt{5}}{5} = \frac{12}{\sqrt{5}}$. Therefore,$	heta = 	an^{-1}(12/\sqrt{5})$.

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 centre of the ellipse $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$ is

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is

The point $(1,3)$ with respect to the ellipse $4x^2+9y^2-16x-54y+61=0$ lies

Let $E$ be the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and $C$ be the circle $x^2 + y^2 = 9$. Let $P$ and $Q$ be the points $(1, 2)$ and $(2, 1)$ respectively. Then

If the line $2x - 3y + 4 = 0$ cuts the ellipse $x = 3 \cos \theta, y = 5 \sin \theta$ at points $A$ and $B$,and $(\alpha, \beta)$ is the midpoint of $\overline{AB}$,then $3\beta - 2\alpha =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module