Question:
The length of the straight line $x - 3y = 1$ intercepted by the hyperbola $x^2 - 4y^2 = 1$ is
The length of the straight line $x - 3y = 1$ intercepted by the hyperbola $x^2 - 4y^2 = 1$ is
Updated On: Aug 19, 2023
- $\sqrt{10} $
- $\frac{6}{5}$
- $ \frac{1}{\sqrt{10}} $
- $\frac{6}{5} \sqrt{10} $
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The Correct Option is D
Solution and Explanation
Given : equation of line, $x - 3y = 1$ (1)
and hyperbola $x^2 - 4y^2 = 1$ (2)
putting $x = 1 + 3y$ in equation (2), we get
$ (1 + 3y)^2 - 4y^2 = 1 \; \Rightarrow \; 1+ 9y^2 + 6y - 4y^2 = 1$
$\Rightarrow \; 5y^2 + 6y = 0 \; \Rightarrow \; y(5y + 6) = 0$
$\Rightarrow $ y = 0 or y = $ - \frac{6}{5}$
$\therefore $ x = 1 for y = 0 & x = 1 - $\frac{18}{5} = \frac{-13}{5}$
for y = -6/5 $\therefore$ the line (1) cuts the hyperbola (2) in at most two point.
$\therefore$ co-ordinates of points are P(1,0) & $Q \left( - \frac{13}{5} , - \frac{6}{5} \right)$
$\therefore \; PQ = \sqrt{ \left(1+ \frac{13}{5}\right)^{2} +\left(0+\frac{6}{5}\right)^{2}} $
$= \sqrt{\left(\frac{18}{5}\right)^{2} +\frac{36}{25}} = \sqrt{\frac{324 +36}{25}} = \sqrt{\frac{360}{25}} $
$ \therefore$ length of straight line $ PQ = \frac{6\sqrt{10}}{5} $ units.
and hyperbola $x^2 - 4y^2 = 1$ (2)
putting $x = 1 + 3y$ in equation (2), we get
$ (1 + 3y)^2 - 4y^2 = 1 \; \Rightarrow \; 1+ 9y^2 + 6y - 4y^2 = 1$
$\Rightarrow \; 5y^2 + 6y = 0 \; \Rightarrow \; y(5y + 6) = 0$
$\Rightarrow $ y = 0 or y = $ - \frac{6}{5}$
$\therefore $ x = 1 for y = 0 & x = 1 - $\frac{18}{5} = \frac{-13}{5}$
for y = -6/5 $\therefore$ the line (1) cuts the hyperbola (2) in at most two point.
$\therefore$ co-ordinates of points are P(1,0) & $Q \left( - \frac{13}{5} , - \frac{6}{5} \right)$
$\therefore \; PQ = \sqrt{ \left(1+ \frac{13}{5}\right)^{2} +\left(0+\frac{6}{5}\right)^{2}} $
$= \sqrt{\left(\frac{18}{5}\right)^{2} +\frac{36}{25}} = \sqrt{\frac{324 +36}{25}} = \sqrt{\frac{360}{25}} $
$ \therefore$ length of straight line $ PQ = \frac{6\sqrt{10}}{5} $ units.
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
