Suppose A, B, C and D are the four intersecti | vedclass

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

Question

(D) Given curves are $\frac{x^2}{18}+\frac{y^2}{8}=1$ $(i)$ and $x^2-y^2=5$ (ii).From (ii), $x^2 = 5+y^2$. Substituting in $(i)$:$\frac{5+y^2}{18} + \

Vedclass · Accepted Answer

$	heta_1=	heta_2=	heta_3=	heta_4$

Vedclass · Answer

(D) Given curves are $\frac{x^2}{18}+\frac{y^2}{8}=1$ $(i)$ and $x^2-y^2=5$ (ii).From (ii), $x^2 = 5+y^2$. Substituting in $(i)$:$\frac{5+y^2}{18} + \frac{y^2}{8} = 1 \Rightarrow \frac{20+4y^2+9y^2}{72} = 1 \Rightarrow 13y^2 = 52 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.If $y = 2$, $x^2 = 9 \Rightarrow x = \pm 3$. If $y = -2$, $x^2 = 9 \Rightarrow x = \pm 3$.Intersection points: $A(3, 2), B(-3, 2), C(-3, -2), D(3, -2)$.Differentiating $(i)$: $\frac{2x}{18} + \frac{2yy'}{8} = 0 \Rightarrow y' = -\frac{4x}{9y}$.Differentiating (ii): $2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y}$.At $A(3, 2)$: $m_1 = -\frac{4(3)}{9(2)} = -\frac{2}{3}$, $m_2 = \frac{3}{2}$.$	an 	heta_1 = |\frac{3/2 - (-2/3)}{1 + (3/2)(-2/3)}| = |\frac{13/6}{0}| 	o \infty \Rightarrow 	heta_1 = 90^\circ$.Due to symmetry of the curves about both axes, the angle of intersection at all four points will be the same.Thus, $	heta_1 = 	heta_2 = 	heta_3 = 	heta_4$.

Suppose $A, B, C$ and $D$ are the four intersection points of the curves $\frac{x^2}{18}+\frac{y^2}{8}=1$ and $x^2-y^2=5$ in $I, II, III$ and $IV$ quadrants respectively. If $\theta_1, \theta_2, \theta_3$ and $\theta_4$ respectively are the angles between the curves at $A, B, C$ and $D$, then

Similar Questions

If $e_{1}$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ where $a > b$,and $e_{2}$ is the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,then find the value of $e_{1}^{2}+e_{2}^{2}$.

If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is $1$,then $b^2=$

$A$ common tangent $T$ to the curves $C_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $T$ touches $C_{1}$ at $(x_{1}, y_{1})$ and $C_{2}$ at $(x_{2}, y_{2})$,then $|2x_{1} + x_{2}|$ is equal to $......$

Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,then the value of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module