If the product of the lengths of the perpendi | vedclass

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

Question

(B) Let the ends of the diameter be $A(2\cos	heta, 2\sin	heta)$ and $B(-2\cos	heta, -2\sin	heta)$.The length of the perpendicular from $(x_1, y_1)

Vedclass · Accepted Answer

$(\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2})$

Vedclass · Answer

(B) Let the ends of the diameter be $A(2\cos	heta, 2\sin	heta)$ and $B(-2\cos	heta, -2\sin	heta)$.The length of the perpendicular from $(x_1, y_1)$ to $x+y+1=0$ is $d = \frac{|x_1+y_1+1|}{\sqrt{1^2+1^2}} = \frac{|x_1+y_1+1|}{\sqrt{2}}$.Let $d_1$ and $d_2$ be the lengths of the perpendiculars from $A$ and $B$ respectively.$d_1 = \frac{|2\cos	heta+2\sin	heta+1|}{\sqrt{2}}$ and $d_2 = \frac{|-2\cos	heta-2\sin	heta+1|}{\sqrt{2}}$.The product $P = d_1 d_2 = \frac{|(1+2(\cos	heta+\sin	heta))(1-2(\cos	heta+\sin	heta))|}{2} = \frac{|1-4(\cos	heta+\sin	heta)^2|}{2}$.$P = \frac{|1-4(1+2\sin	heta\cos	heta)|}{2} = \frac{|1-4-4\sin(2	heta)|}{2} = \frac{|-3-4\sin(2	heta)|}{2} = \frac{3+4\sin(2	heta)}{2}$.For $P$ to be maximum,$\sin(2	heta) = 1$,which means $2	heta = \frac{\pi}{2}$,so $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$ into the coordinates of $A$ and $B$:$A = (2\cos\frac{\pi}{4}, 2\sin\frac{\pi}{4}) = (2\frac{1}{\sqrt{2}}, 2\frac{1}{\sqrt{2}}) = (\sqrt{2}, \sqrt{2})$.$B = (-2\cos\frac{\pi}{4}, -2\sin\frac{\pi}{4}) = (-\sqrt{2}, -\sqrt{2})$.Thus,the correct option is $B$.

If the product of the lengths of the perpendiculars drawn from the ends of a diameter of the circle $x^2+y^2=4$ onto the line $x+y+1=0$ is maximum,then the two ends of that diameter are

Similar Questions

Let $A, B, C$ be three points on a circle of radius $1$ such that $\angle ACB = \frac{\pi}{4}$. Then,the length of the side $AB$ is

If the lines joining the origin to the points of intersection of a line $L$ and $x^2+y^2=4$ are the coordinate axes,then the equation of the line $L$ is

If $C(\alpha, \beta)$ with $\alpha < 0$ is the centre of the circle that touches the $Y$-axis at $(0, 3)$ and makes an intercept of length $2$ units on the positive $X$-axis,then $(\alpha, \beta) =$

The angle subtended at the centre of a circle of radius $3 \text{ metres}$ by an arc of length $1 \text{ metre}$ is equal to

The set of values of $k$ for which the circle $C : 4x^{2} + 4y^{2} - 12x + 8y + k = 0$ lies inside the fourth quadrant and the point $(1, -1/3)$ lies on or inside the circle $C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module