If 2p is the length of the perpendicular from | vedclass

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

Question

(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The length of the perpendicular from the origin $(0,0)$ to the line is given by $d

Vedclass · Accepted Answer

$a^2, 8p^2, b^2$ are in $H.P.$

Vedclass · Answer

(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The length of the perpendicular from the origin $(0,0)$ to the line is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Here,$2p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,$4p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.Multiplying by $2$,we get $\frac{2}{a^2} + \frac{2}{b^2} = \frac{2}{4p^2} = \frac{1}{2p^2}$.Wait,the standard relation is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$ where $p$ is the perpendicular. Given $2p$ is the length,let $L = 2p$. Then $\frac{1}{L^2} = \frac{1}{a^2} + \frac{1}{b^2}$.So,$\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.This means $\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ if we consider the harmonic mean property,but specifically,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.Actually,$\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ implies $\frac{2}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{2p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.Given $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,we have $\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ in $A.P.$ implies $\frac{2}{8p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.Thus,$\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ are in $A.P.$,which implies $a^2, 8p^2, b^2$ are in $H.P.$

If $2p$ is the length of the perpendicular from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$,then:

Similar Questions

What are the points on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units?

Let the centroid of an equilateral triangle $ABC$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x + y = 3$. If $R$ and $r$ are the radius of the circumcircle and incircle respectively of $\Delta ABC$,then $(R + r)$ is equal to ..... .

Two sides of a square are along the lines $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$. The area of the square is:

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

Two points $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$. If $P$ is the point of intersection of the lines $4x + 3y + k = 0$ and $3x + 4y + k = 0$,then the range of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module