If a circle cuts a rectangular hyperbola xy = | vedclass

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

Question

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + k = 0$.The parametric coordinates of any point on the rectangular hyperbola $xy = c^2$

Vedclass · Accepted Answer

$t_1t_2t_3t_4 = 1$

Vedclass · Answer

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + k = 0$.The parametric coordinates of any point on the rectangular hyperbola $xy = c^2$ are $(ct, c/t)$.Substituting these into the circle equation:$(ct)^2 + (c/t)^2 + 2g(ct) + 2f(c/t) + k = 0$Multiplying by $t^2$ to clear the denominator:$c^2t^4 + 2gct^3 + kt^2 + 2fct + c^2 = 0$This is a biquadratic equation in $t$ whose roots are $t_1, t_2, t_3,$ and $t_4$.From the properties of roots of a polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$,the product of the roots is given by $(-1)^n \frac{a_0}{a_n}$.Here,the product of the roots is $t_1t_2t_3t_4 = (-1)^4 \frac{c^2}{c^2} = 1$.

If a circle cuts a rectangular hyperbola $xy = c^2$ at four points $A, B, C,$ and $D$,and the parameters of these four points are $t_1, t_2, t_3,$ and $t_4$ respectively,then which of the following is true?

Similar Questions

Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is

The Cartesian form of the curve given by $x = \frac{a}{2} (t + \frac{1}{t})$ and $y = \frac{a}{2} (t - \frac{1}{t})$,where $t$ is a parameter,is:

The equation of the hyperbola having its eccentricity $e = 2$ and the distance between its foci as $8$ is:

If $3x + 2\sqrt{2}y + k = 0$ is a normal to the hyperbola $4x^2 - 9y^2 - 36 = 0$ making positive intercepts on both the axes,then $k=$ (in $\sqrt{2}$)

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}t = 0$ and $\sqrt{3}tx + ty - 4\sqrt{3} = 0$ (where $t$ is a parameter) is a hyperbola whose eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module