Let a circle of radius 4 and the ellipse 15x^ | vedclass

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

Question

(B) The equation of the ellipse is $15x^2 + 19y^2 = 285$,which can be written as $\frac{x^2}{19} + \frac{y^2}{15} = 1$.Here,$a^2 = 19$ and $b^2 = 15$.

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(B) The equation of the ellipse is $15x^2 + 19y^2 = 285$,which can be written as $\frac{x^2}{19} + \frac{y^2}{15} = 1$.Here,$a^2 = 19$ and $b^2 = 15$.The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{19m^2 + 15}$.This line is also a tangent to the circle $x^2 + y^2 = 4^2 = 16$,so its perpendicular distance from the center $(0,0)$ must be equal to the radius $4$.$\left| \frac{\pm \sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} ight| = 4$.Squaring both sides,we get $\frac{19m^2 + 15}{m^2 + 1} = 16$.$19m^2 + 15 = 16m^2 + 16$,which simplifies to $3m^2 = 1$,so $m = \pm \frac{1}{\sqrt{3}}$.The angle $	heta$ that the tangent makes with the $x$-axis is given by $	an 	heta = |m| = \frac{1}{\sqrt{3}}$,so $	heta = \frac{\pi}{6}$.The minor axis of the ellipse is the $y$-axis (since $b^2 The angle between the tangent and the $y$-axis is $\frac{\pi}{2} - 	heta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

Let a circle of radius $4$ and the ellipse $15x^2 + 19y^2 = 285$ be concentric. What is the angle that the common tangents make with the minor axis of the ellipse?

Similar Questions

Which of the following lines is a tangent to the circle $x^2 + y^2 = 25$ for all values of $m$?

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$,then its radius is

The point on the curve ${y^2} = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ is

The equation of the pair of tangents drawn from the point $(1, 1)$ to the circle $x^2 + y^2 + 2x + 2y + 1 = 0$ is:

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at $P(-4,0)$. Then the area of quadrilateral $PAOB$,where $O$ is the origin,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module