Consider a rigid square ABCD as in the figure | vedclass

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

Question

(D) Let the side length of the square $ABCD$ be $a$. Let the coordinates of $A$ be $(x_A, 0)$ and $B$ be $(0, y_B)$. Let the angle $\angle OAB = 	het

Vedclass · Accepted Answer

an ellipse which is not a circle

Vedclass · Answer

(D) Let the side length of the square $ABCD$ be $a$. Let the coordinates of $A$ be $(x_A, 0)$ and $B$ be $(0, y_B)$. Let the angle $\angle OAB = 	heta$. Then $A = (a \cos 	heta, 0)$ and $B = (0, a \sin 	heta)$.Since $ABCD$ is a square,the vector $\vec{BC}$ is perpendicular to $\vec{AB}$ and has the same length $a$. The vector $\vec{AB} = (-a \cos 	heta, a \sin 	heta)$. Rotating this by $90^\circ$ gives $\vec{BC} = (a \sin 	heta, a \cos 	heta)$.Thus,$C = B + \vec{BC} = (0 + a \sin 	heta, a \sin 	heta + a \cos 	heta) = (a \sin 	heta, a(\sin 	heta + \cos 	heta))$.Let $C = (x, y)$. Then $x = a \sin 	heta$ and $y = a \sin 	heta + a \cos 	heta$.From the first equation,$\sin 	heta = \frac{x}{a}$. Then $\cos 	heta = \sqrt{1 - (\frac{x}{a})^2} = \frac{\sqrt{a^2 - x^2}}{a}$.Substituting into the second equation: $y = x + \sqrt{a^2 - x^2}$.$y - x = \sqrt{a^2 - x^2}$.Squaring both sides: $(y - x)^2 = a^2 - x^2$.$y^2 - 2xy + x^2 = a^2 - x^2$.$2x^2 + y^2 - 2xy = a^2$.This is the equation of an ellipse,as the discriminant $B^2 - 4AC = (-2)^2 - 4(2)(1) = 4 - 8 = -4 < 0$. Since the coefficients of $x^2$ and $y^2$ are not equal,it is not a circle.

Consider a rigid square $ABCD$ as in the figure with $A$ and $B$ on the $X$ and $Y$-axes,respectively. When $A$ and $B$ slide along their respective axes,the locus of $C$ forms a part of

Similar Questions

If the line $lx + my = 1$ is a tangent to the circle ${x^2} + {y^2} = {a^2}$,then the locus of the point $(l, m)$ is

The equation of the director circle of the circle ${x^2} + {y^2} = {a^2}$ is

The locus of the center of the circle which cuts the circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ orthogonally is

If a system of circles passes through $(2,3)$ and cuts the circle $x^2+y^2=12$ orthogonally,then the equation of the locus of the centres of that system of circles is:

The tangent at $P$,any point on the circle ${x^2} + {y^2} = 4$,meets the coordinate axes in $A$ and $B$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module