If the circle x^2+y^2=a^2 intersects the hype | vedclass

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

Question

(A) Given equations are $x^2+y^2=a^2$ and $xy=c^2$.From the second equation,$x = \frac{c^2}{y}$.Substituting $x$ in the first equation:$\left(\frac{c^

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given equations are $x^2+y^2=a^2$ and $xy=c^2$.From the second equation,$x = \frac{c^2}{y}$.Substituting $x$ in the first equation:$\left(\frac{c^2}{y}\right)^2 + y^2 = a^2$$\frac{c^4}{y^2} + y^2 = a^2$$c^4 + y^4 = a^2 y^2$$y^4 - a^2 y^2 + c^4 = 0$This is a biquadratic equation in $y$. Let the roots be $y_1, y_2, y_3, y_4$.The equation is of the form $Ay^4 + By^3 + Cy^2 + Dy + E = 0$,where $B=0$.By Vieta's formulas,the sum of the roots is $-\frac{B}{A} = -\frac{0}{1} = 0$.Therefore,$y_1+y_2+y_3+y_4 = 0$.

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $(x_i, y_i)$,for $i=1, 2, 3, 4$,then $y_1+y_2+y_3+y_4$ equals

Similar Questions

The foci of the hyperbola $2x^2 - 3y^2 = 5$ are

The equation of a tangent to the hyperbola $16x^2 - 25y^2 - 96x + 100y - 356 = 0$ which makes an angle $45^{\circ}$ with its transverse axis is

The equation ${x^2} - 16xy - 11{y^2} - 12x + 6y + 21 = 0$ represents

$A$ hyperbola,having the transverse axis of length $2 \sin \theta$,is confocal with the ellipse $3 x^2 + 4 y^2 = 12$. Then its equation is

If the product of the slopes of the tangents drawn from an external point $P(h, k)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a constant $k^2$,then the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module