If the circle x^{2}+y^{2}=a^{2} intersects th | vedclass

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

Question

(D) Given equations are $x^{2}+y^{2}=a^{2}$ and $xy=c^{2}$.Substituting $y = \frac{c^{2}}{x}$ into the circle equation:$x^{2} + (\frac{c^{2}}{x})^{2}

Vedclass · Accepted Answer

$x_{1}+x_{2}+x_{3}+x_{4}=0$

Vedclass · Answer

(D) Given equations are $x^{2}+y^{2}=a^{2}$ and $xy=c^{2}$.Substituting $y = \frac{c^{2}}{x}$ into the circle equation:$x^{2} + (\frac{c^{2}}{x})^{2} = a^{2}$$x^{2} + \frac{c^{4}}{x^{2}} = a^{2}$$x^{4} - a^{2}x^{2} + c^{4} = 0$This is a biquadratic equation in $x$. The roots are $x_{1}, x_{2}, x_{3}, x_{4}$.Comparing this with the standard form $Ax^{4} + Bx^{3} + Cx^{2} + Dx + E = 0$,we have $B=0$.By Vieta's formulas,the sum of the roots $x_{1}+x_{2}+x_{3}+x_{4} = -\frac{B}{A} = 0$.

If the circle $x^{2}+y^{2}=a^{2}$ intersects the hyperbola $xy=c^{2}$ in four points $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3})$ and $S(x_{4}, y_{4})$,then

Similar Questions

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=b^2$ at four points $(x_1, y_1)$,$(x_2, y_2)$,$(x_3, y_3)$,and $(x_4, y_4)$,then $y_1 y_2 y_3 y_4 = $

The area of the quadrilateral formed by the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units):

Find the equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$.

The equation of the normal at the point $(a \sec \theta, b \tan \theta)$ of the curve $b^2 x^2 - a^2 y^2 = a^2 b^2$ is

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 3 \sqrt{5}, 0)$,the latus rectum is of length $8$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module