First slide

Pair of straight lines

Question

The Locus of the midpoint of the portion of the line $3 xsecθ + 4 ytanθ = 1$ intercepted between the axes is $\frac{1}{{ax}^{2}} - \frac{1}{{by}^{2}} = 1$ then $a + b =$

Moderate

A

20
B

30
C

50
D

100

Solution

The equation of the line whose portion of the line intercepted between the axes is bisected by the point $P (x_{1}, y_{1})$ is $\frac{x}{2 x_{1}} + \frac{y}{2 y_{1}} = 1$
Suppose that $P (x_{1}, y_{1})$ be any point on the locus.
Compare the equation $3 xsecθ + 4 ytanθ = 1$ with $\frac{x}{2 x_{1}} + \frac{y}{2 y_{1}} = 1$
It implies
$\begin{matrix} 3 secθ = \frac{1}{2 x_{1}} \\ secθ = \frac{1}{6 x_{1}} \end{matrix}$
And
$\begin{matrix} 4 \tan θ = \frac{1}{2 y_{1}} \\ \tan θ = \frac{1}{8 y_{1}} \end{matrix}$
To eliminate $θ$ , use the trigonometric identity $\sec^{2} θ - \tan^{2} θ = 1$
It implies that $\frac{1}{36 {x_{1}}^{2}} - \frac{1}{64 {y_{1}}^{2}} = 1$

Therefore, the required locus is $\frac{1}{36 x^{2}} - \frac{1}{64 y^{2}} = 1$ , $a + b = 100$

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