Let L_1 be the length of the common chord of | vedclass

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

Question

(C) The equations of the curves are $x^2 + y^2 = 9$ and $y^2 = 8x$.Substituting $y^2 = 8x$ into the circle equation: $x^2 + 8x = 9$.$x^2 + 8x - 9 = 0$

Vedclass · Accepted Answer

$L_1 < L_2$

Vedclass · Answer

(C) The equations of the curves are $x^2 + y^2 = 9$ and $y^2 = 8x$.Substituting $y^2 = 8x$ into the circle equation: $x^2 + 8x = 9$.$x^2 + 8x - 9 = 0$.$(x + 9)(x - 1) = 0$.Since $x$ must be non-negative for the parabola $y^2 = 8x$,we have $x = 1$.For $x = 1$,$y^2 = 8(1) = 8$,so $y = \pm 2\sqrt{2}$.The points of intersection are $A(1, 2\sqrt{2})$ and $B(1, -2\sqrt{2})$.The length of the common chord $L_1 = |2\sqrt{2} - (-2\sqrt{2})| = 4\sqrt{2} \approx 5.66$.The parabola $y^2 = 8x$ is of the form $y^2 = 4ax$,where $4a = 8$,so $a = 2$.The length of the latus rectum $L_2 = 4a = 8$.Comparing the values,$4\sqrt{2} < 8$,therefore $L_1 < L_2$.

Let $L_1$ be the length of the common chord of the curves $x^2 + y^2 = 9$ and $y^2 = 8x$,and $L_2$ be the length of the latus rectum of $y^2 = 8x$,then

Similar Questions

The equation of the pair of tangents at $(0,1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$ is

$y = mx$ is a chord of a circle of radius $a$. The diameter of the circle lies along the $x$-axis and one end of this chord is at the origin. The equation of the circle described on this chord as diameter is

The length of the common chord of the circles $(x - a)^{2} + y^{2} = c^{2}$ and $x^{2} + (y - b)^{2} = c^{2}$ is .....

If the circle $S \equiv x^2+y^2-4=0$ intersects another circle $S^{\prime}=0$ of radius $\frac{5 \sqrt{2}}{2}$ in such a manner that the common chord is of maximum length with slope equal to $\frac{1}{4}$,then the centre of $S^{\prime}=0$ is

The length of the common chord of the circles ${x^2} + {y^2} + 5x + 7y + 9 = 0$ and ${x^2} + {y^2} + 7x + 5y + 9 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module