Let θ be the acute angle between the tangents to the ellipse (x2/9)+(y2/1)=1 and the circle x2+y2=3 at their point of intersection in the first quadrant. Then tan θ is equal to :

Question

Let θ be the acute angle between the tangents to the ellipse (x2/9)+(y2/1)=1 and the circle x2+y2=3 at their point of intersection in the first quadrant. Then tan θ is equal to : (A) (5/2 √3) (B) (2/√3) (C) (4/√3) (D) 2

Tardigrade Admin · Accepted Answer

The point of intersection of the curves

\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1

and

x^{2} + y^{2} = 3

in the first quadrant is

(\frac{3}{2}, \frac{3}{2})

Now slope of tangent to the ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1

at

(\frac{3}{2}, \frac{3}{2})

is

m_{1} = - \frac{1}{3 3}

And slope of tangent to the circle at

(\frac{3}{2}, \frac{3}{2})

is

m_{2}

= - 3

So, if angle between both curves is

θ

then

tan θ = ∣ ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣ ∣ = ∣ ∣ \frac{- \frac{1}{3 3} + 3}{1 + ( - \frac{1}{3 3} ( - 3 ) )} ∣ ∣

= \frac{2}{3}

Q. Let $θ$ be the acute angle between the tangents to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1$ and the circle $x^{2} + y^{2} = 3$ at their point of intersection in the first quadrant. Then $tan θ$ is equal to :

$\frac{5}{2 3}$

$\frac{2}{3}$

$\frac{4}{3}$

$2$

Q. Let θ be the acute angle between the tangents to the ellipse 9x2​+1y2​=1 and the circle x2+y2=3 at their point of intersection in the first quadrant. Then tanθ is equal to :

23​5​

3​2​

3​4​

2

Q. Let $θ$ be the acute angle between the tangents to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1$ and the circle $x^{2} + y^{2} = 3$ at their point of intersection in the first quadrant. Then $tan θ$ is equal to :

$\frac{5}{2 3}$

$\frac{2}{3}$

$\frac{4}{3}$

$2$