The product of the perpendiculars drawn from | vedclass

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

Question

(B) The equation of the line is $\frac{x}{a}\cos 	heta + \frac{y}{b}\sin 	heta - 1 = 0$,which can be written as $bx \cos 	heta + ay \sin 	heta - a

Vedclass · Accepted Answer

$b^2$

Vedclass · Answer

(B) The equation of the line is $\frac{x}{a}\cos 	heta + \frac{y}{b}\sin 	heta - 1 = 0$,which can be written as $bx \cos 	heta + ay \sin 	heta - ab = 0$.The perpendicular distance $p_1$ from $(\sqrt{a^2 - b^2}, 0)$ is $p_1 = \frac{|b(\sqrt{a^2 - b^2})\cos 	heta + a(0)\sin 	heta - ab|}{\sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}} = \frac{|b\sqrt{a^2 - b^2}\cos 	heta - ab|}{\sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}}$.The perpendicular distance $p_2$ from $(-\sqrt{a^2 - b^2}, 0)$ is $p_2 = \frac{|b(-\sqrt{a^2 - b^2})\cos 	heta + a(0)\sin 	heta - ab|}{\sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}} = \frac{|-b\sqrt{a^2 - b^2}\cos 	heta - ab|}{\sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}}$.The product $p_1 p_2 = \frac{|(b\sqrt{a^2 - b^2}\cos 	heta - ab)(-b\sqrt{a^2 - b^2}\cos 	heta - ab)|}{b^2 \cos^2 	heta + a^2 \sin^2 	heta}$.Using $(x-y)(x+y) = x^2 - y^2$,the numerator becomes $|(ab)^2 - (b\sqrt{a^2 - b^2}\cos 	heta)^2| = |a^2 b^2 - b^2(a^2 - b^2)\cos^2 	heta|$.$p_1 p_2 = \frac{b^2 |a^2 - (a^2 - b^2)\cos^2 	heta|}{b^2 \cos^2 	heta + a^2 \sin^2 	heta} = \frac{b^2 |a^2 - a^2 \cos^2 	heta + b^2 \cos^2 	heta|}{b^2 \cos^2 	heta + a^2 \sin^2 	heta} = \frac{b^2 |a^2 \sin^2 	heta + b^2 \cos^2 	heta|}{b^2 \cos^2 	heta + a^2 \sin^2 	heta} = b^2$.

The product of the perpendiculars drawn from the points $(\pm \sqrt{a^2 - b^2}, 0)$ to the line $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$ is:

Similar Questions

An ellipse has $6$ and $2$ as the lengths of its major and minor axes,respectively. If the center is at $(5,6)$ and the major axis is along $x-y+1=0$,then the equation of the ellipse is

At which points on the ellipse $4x^2 + 9y^2 = 1$ is the tangent line parallel to the line $8x = 9y$?

The number of values of $c$ such that the line $y = cx + c$,where $c \in R$,touches the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is:

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$.

The length of the latus rectum of $16x^2 + 25y^2 = 400$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module