If θ is the acute angle of intersection at a real point of intersection of the circle x2 + y2 = 5 and the parabola y 2= 4x then Tan θ is equal to

Question

If θ is the acute angle of intersection at a real point of intersection of the circle x2 + y2 = 5 and the parabola y 2= 4x then Tan θ is equal to (A) 1 (B) √ 3 (C) 3 (D) (1/√3)

Tardigrade Admin · Accepted Answer

Given equations are x2+y2=5 dots(i) and y2=4 x dots(ii) On solving Eq. (i) and (ii), we get x=-5,1 at x=-5, y2=-20 (imaginary value) ∴ at x=1, y2=4 ⇒ y=± 2 Hence, point of intersection are (1,2) and (1,-2) On differentiating Eq. (i) w. r. t. x, we get 2 x+2 y (d y/d x)=0 ⇒ (d y/d x)=-(x/y) ∴ m1=((d y/d x))(1,2)=-(1/2) And on differentiating Eq. (ii) w. r. t. x, we get 2 y (d y/d x)=4 ⇒ (d y/d x)=(2/y) m2=((d y/d x))(1,2)=(2/2)=1 Now, tan θ =|(m1-m2/1+m1 m2)| =|(-(1/2)-1/1-(1)2)|=|(-(3/2)/(1)2)|=3

Q. If $θ$ is the acute angle of intersection at a real point of intersection of the circle $x^{2} + y^{2} = 5$ and the parabola $y^{2} = 4 x$ then Tan $θ$ is equal to

$1$

$3$

$3$

$\frac{1}{3}$

Q. If θ is the acute angle of intersection at a real point of intersection of the circle x2+y2=5 and the parabola y2=4x then Tan θ is equal to

1

3​

3

3​1​

Q. If $θ$ is the acute angle of intersection at a real point of intersection of the circle $x^{2} + y^{2} = 5$ and the parabola $y^{2} = 4 x$ then Tan $θ$ is equal to

$1$

$3$

$3$

$\frac{1}{3}$