First slide

Pair of straight lines

Question

The product of the perpendicular from (1,1) to the pair of lines ${ax}^{2} - 2 hxy + {by}^{2} = 0$ is 1 then $\frac{1}{a} + \frac{1}{b}$

Moderate

A

$\frac{1}{h}$
B

$\frac{3}{h}$
C

2h
D

4h

Solution

The product of the perpendiculars drawn from the point $P (x_{1}, y_{1})$ to the pair of lines represented by the equation ${ax}^{2} + 2 hxy + {by}^{2} = 0$ is $|\frac{{ax}_{1}^{2} + 2 {hx}_{1} y_{1} + {by}_{1}^{2}}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}|$
For the equation ${ax}^{2} - 2 hxy + {by}^{2} = 0$ the product of the perpendiculars from the point (1,1) is 1
$\begin{array}{l} 1 = |\frac{a {(1)}^{2} - 2 h (1) (1) + b {(1)}^{2}}{\sqrt{{(a - b)}^{2} + {(2 h)}^{2}}}| \\ 1 = |\frac{a - 2 h + b}{\sqrt{{(a - b)}^{2} + 4 h^{2}}}| \end{array}$
Squaring on both sides and then simplify,
$\begin{matrix} {(a + b - 2 h)}^{2} = {(a - b)}^{2} + 4 h^{2} \\ {(a + b)}^{2} + 4 h^{2} - 4 h (a + b) = {(a - b)}^{2} + 4 h^{2} \\ {(a + b)}^{2} - {(a - b)}^{2} = 4 h (a + b) \\ 4 ab = 4 h (a + b) \end{matrix}$
Therefore,
$\frac{1}{h} = \frac{1}{a} + \frac{1}{b}$
Therefore, the value of $\frac{1}{a} + \frac{1}{b}$ is $\frac{1}{h}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App