A line L makes intercepts a and b on the coor | vedclass

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

Question

(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.The perpendicular distance $d$ from the origi

Vedclass · Accepted Answer

$\frac{1}{p^2}+\frac{1}{q^2}$

Vedclass · Answer

(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by $d = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,we get $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2}$.When the axes are rotated,the origin remains fixed,so the perpendicular distance $d$ from the origin to the line $L$ remains invariant.In the new coordinate system,the line $L$ makes intercepts $p$ and $q$,so its equation is $\frac{x}{p} + \frac{y}{q} = 1$.The perpendicular distance $d$ from the origin to this new line is $d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}$,which implies $\frac{1}{d^2} = \frac{1}{p^2} + \frac{1}{q^2}$.Since $d$ is the same in both cases,we have $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

$A$ line $L$ makes intercepts $a$ and $b$ on the coordinate axes. The axes are rotated through an angle $\theta$ in the positive direction,keeping the origin fixed. If the line $L$ makes intercepts $p$ and $q$ on the new coordinate axes,then $\frac{1}{a^2}+\frac{1}{b^2}=$

Similar Questions

When the coordinate axes are rotated through an angle $135^{\circ}$,the coordinates of a point $P$ in the new system are known to be $(4, -3)$. Find the coordinates of $P$ in the original system.

When the origin is shifted to $(h, k)$ by translation of axes,the transformed equation of $x^2+2x+2y-7=0$ does not contain the $x$ term and the constant term. Then $(2h+k) =$

The transformed equation of $x^2+y^2=r^2$ when the axes are rotated through an angle $36^{\circ}$ is

If the coordinate axes are rotated through an angle $\frac{\pi}{6}$ about the origin,then the transformed equation of $\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$ is

The point to which the origin should be shifted so that the equation $y^2-6y-4x+13=0$ will not contain any term in $y$ and the constant term,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module