The product of perpendicular drawn from the origin to the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , will be:

\frac{a b}{\sqrt{a^{2} - b^{2} + 4 h^{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{b c}{\sqrt{a^{2} - b^{2} + 4 h^{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c a}{\sqrt{a^{2} - b^{2} + 4 h^{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c}{\sqrt{(a - b)^{2} + 4 h^{2}}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{c}{\sqrt{(a - b)^{2} + 4 h^{2}}}$
Let $l_{1} x + m_{1} y + n_{1} = 0$ & $l_{2} x + m_{2} y + n_{2} = θ$ be two lines represented by ${a x}^{2} + 2 h x y + {b y}^{2} + 2 g x + 2 f y + c = 0$
$l_{1} l_{2} = a; (l_{1} m_{2} + l_{2} n_{1}) = 2 h$ , $m_{1} m_{2} = b, (l_{1} n_{2} + l_{2} n_{1}) = 2 g, (m_{1} n_{2} + m_{2} n_{1}) = 2 f$
$n_{1} n_{2} = c$
Perpendicular from origin to $l_{1} x + m_{1} y + n_{1} = 0$
$= \frac{n_{1}}{\sqrt{l_{1}^{2} + m_{1}^{2}}}$
Perpendicular from origin to $l_{2} x + m_{2} y + n_{2} = 0$
$= \frac{n_{2}}{\sqrt{l_{2}^{2} + m_{2}^{2}}}$
Product, $\frac{n_{1} n_{2}}{\sqrt{(l_{1}^{2} + m_{1}^{2}) (l_{2}^{2} + m_{2}^{2})}}$
$= \frac{C}{\sqrt{{(l_{1} l_{2})}^{2} + (l_{1}^{2} m_{2}^{2} + l_{2}^{2} m_{1}^{2}) + {(m_{1} m_{2})}^{2}}}$
$= \frac{C}{\sqrt{{(l_{1} l_{2})}^{2} + {(l_{1} m_{2} + l_{2} m_{1})}^{2} {- 2 l_{1} l_{2} m_{1} m_{2} + (m_{1} m_{2})}^{2}}}$
$= \frac{C}{\sqrt{a^{2} + {4 h}^{2} - 2 a b + b^{2}}}$
$= \frac{C}{\sqrt{{(a - b)}^{2} + {4 h}^{2}}}$

Suggest Corrections

Similar questions

The product of perpendiculars drawn from the origin to the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0,$ will be

Q. Assertion :If the equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represent two straight lines, then product of perpendiculars drawn from the origin to these lines is

∣ ∣ ∣ ∣ ∣ \frac{c}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} ∣ ∣ ∣ ∣ ∣

Reason: The perpendicular distance from origin to line

A x + B y + k = 0

\frac{| k |}{\sqrt{A^{2} + B^{2}}}

Q. Prove that the squares of the semi-axes of the conic

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

are

2 △ \div {(a b - h^{2}) (a + b \pm \sqrt{(a - b)^{2} + 4 h^{2}})},

where

△

is the discriminant.

Q. Prove that the product the lengths of perpendicular form

P (x_{1}, y_{1})

to the line represented

{a x}^{2} + 2 h x y + {b y}^{2} = 0

\frac{{a x}_{1}^{2} + 2 {h x}_{1} y_{1} + {b y}_{1}^{2}}{\sqrt{{(a - b)}^{2} + {4 h}^{2}}}

If the equation $a x^{2} + 2 h x y + b y^{2} = 0$ has one line as the bisector of angle between the coordinate axes, then

Join BYJU'S Learning Program

The product of perpendicular drawn from the origin to the lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0, will be:

The product of perpendicular drawn from the origin to the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , will be: