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

$A$ particle is projected vertically upwards. If it has to stay above the ground for $12 \text{ seconds}$, then:

The locus of a variable point which forms a triangle of fixed area with two fixed points is

Let $A$ be the point $(0,4)$ and $B$ be a moving point on the $x$-axis. Let $M$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the midpoint $P$ of $MR$ is

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,find its locus.

If a point $(x, y) = (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module