A moving line intersects the lines x+y=0 and | vedclass

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

Question

(B) Let the mid-point of $AB$ be $(h, k)$.Let $A = (\alpha, -\alpha)$ and $B = (\beta, \beta)$.Then, the mid-point $(h, k) = \left(\frac{\alpha+\beta}

Vedclass · Accepted Answer

$\left(x^{2}-y^{2}\right)^{2}=C^{2}$

Vedclass · Answer

(B) Let the mid-point of $AB$ be $(h, k)$.Let $A = (\alpha, -\alpha)$ and $B = (\beta, \beta)$.Then, the mid-point $(h, k) = \left(\frac{\alpha+\beta}{2}, \frac{\beta-\alpha}{2}ight)$.So, $\alpha+\beta = 2h$ and $\beta-\alpha = 2k$.The area of $	riangle AOB = \frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} \sqrt{\alpha^{2} + (-\alpha)^{2}} \sqrt{\beta^{2} + \beta^{2}} = \frac{1}{2} \sqrt{2\alpha^{2}} \sqrt{2\beta^{2}} = |\alpha\beta| = C$.Thus, $\alpha^{2}\beta^{2} = C^{2}$.We know that $(\beta+\alpha)^{2} - (\beta-\alpha)^{2} = 4\alpha\beta$.So, $4\alpha\beta = (2h)^{2} - (2k)^{2} = 4(h^{2}-k^{2})$, which implies $\alpha\beta = h^{2}-k^{2}$.Substituting this into $\alpha^{2}\beta^{2} = C^{2}$, we get $(h^{2}-k^{2})^{2} = C^{2}$.Replacing $(h, k)$ with $(x, y)$, the locus is $(x^{2}-y^{2})^{2} = C^{2}$.

$A$ moving line intersects the lines $x+y=0$ and $x-y=0$ at the points $A$ and $B$ respectively, such that the area of the triangle with vertices $(0,0)$, $A$, and $B$ has a constant area $C$. The locus of the mid-point of $AB$ is given by the equation:

Similar Questions

If the algebraic sum of the perpendicular distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ to a variable line is zero,then the variable line always passes through a fixed point. The coordinates of that point are

The coordinates of a point $P$ on the line $2x - y + 5 = 0$ such that $|PA - PB|$ is maximum,where $A$ is $(4, -2)$ and $B$ is $(2, -4)$,are:

Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $PA^2+PB^2=2PC^2$,then a point that lies on the locus of $P$ is

$A$ straight rod of length $4$ units slides such that its ends $A$ and $B$ always lie on the $X$ and $Y$-axes respectively. Then,the locus of the centroid of $\triangle OAB$ is

If the line $y = \sqrt{3}x$ cuts the curve $x^3 + 3y^2 + 4x + 5y - 1 = 0$ at the points $A, B, C$,then the product $OA \cdot OB \cdot OC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module