The angle of intersection between the curves | vedclass

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

Question

(B) Given curves are $y^2+x^2=a^2 \sqrt{2}$ $(i)$ and $x^2-y^2=a^2$ $(ii)$.Adding $(i)$ and $(ii)$,we get $2x^2 = a^2(\sqrt{2}+1) \Rightarrow x^2 = \f

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Given curves are $y^2+x^2=a^2 \sqrt{2}$ $(i)$ and $x^2-y^2=a^2$ $(ii)$.Adding $(i)$ and $(ii)$,we get $2x^2 = a^2(\sqrt{2}+1) \Rightarrow x^2 = \frac{a^2(\sqrt{2}+1)}{2}$.Subtracting $(ii)$ from $(i)$,we get $2y^2 = a^2(\sqrt{2}-1) \Rightarrow y^2 = \frac{a^2(\sqrt{2}-1)}{2}$.For curve $(i)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{dy}{dx} = -\frac{x}{y}$.For curve $(ii)$,differentiating with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{dy}{dx} = \frac{x}{y}$.The angle of intersection $	heta$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.Substituting $m_1 = -\frac{x}{y}$ and $m_2 = \frac{x}{y}$:$	an 	heta = |\frac{-\frac{x}{y} - \frac{x}{y}}{1 + (-\frac{x}{y})(\frac{x}{y})}| = |\frac{-2x/y}{1 - x^2/y^2}| = |\frac{-2xy}{y^2 - x^2}|$.Since $x^2 = \frac{a^2(\sqrt{2}+1)}{2}$ and $y^2 = \frac{a^2(\sqrt{2}-1)}{2}$,then $y^2 - x^2 = -a^2$.Also $x^2 y^2 = \frac{a^4(2-1)}{4} = \frac{a^4}{4} \Rightarrow xy = \frac{a^2}{2}$.$	an 	heta = |\frac{-2(a^2/2)}{-a^2}| = |\frac{-a^2}{-a^2}| = 1$.Therefore,$	heta = \frac{\pi}{4}$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

The angle of intersection between the curves $y^2+x^2=a^2 \sqrt{2}$ and $x^2-y^2=a^2$ is

Similar Questions

Each of the two orthogonal circles $C_1$ and $C_2$ passes through both the points $(2,0)$ and $(-2,0)$. If $y=mx+c$ is a common tangent to these circles,then

Prove that the curves $y^{2}=4x$ and $x^{2}+y^{2}-6x+1=0$ touch each other at the point $(1,2)$.

If $\theta$ is the angle subtended at $P(x_1, y_1)$ by the circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$,then

If the equation of the circumcircle of the triangle formed by the lines $L_1 \equiv x+y=0$,$L_2 \equiv 2x+y-1=0$,and $L_3 \equiv x-3y+2=0$ is $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$,then find the value of $\frac{7 \lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module