Find the value of k,if the angle between the | vedclass

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

Question

(C) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4

Vedclass · Accepted Answer

$\pm \frac{1}{5}$

Vedclass · Answer

(C) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$,we get $a = 2$,$h = \frac{5}{2}$,and $b = 3$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(\frac{5}{2})^2 - (2)(3)}}{2 + 3} ight|$.$	an 	heta = \left| \frac{2\sqrt{\frac{25}{4} - 6}}{5} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{5} ight| = \left| \frac{2 	imes \frac{1}{2}}{5} ight| = \frac{1}{5}$.Since the angle between the lines is $	an^{-1}(k)$,we have $	an 	heta = |k| = \frac{1}{5}$,which implies $k = \pm \frac{1}{5}$.

Find the value of $k$,if the angle between the straight lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1}(k)$.

Similar Questions

The formula of sodium nitroprusside is:

We consider the radiation emitted by the human body. Which of the following statements is true?

The least count of a stopwatch is $\frac{1}{5} \, s$. The time of $20$ oscillations of a pendulum is measured to be $25 \, s$. The maximum percentage error in the measurement of time will be ..... $\%$

Which of the following statements is incorrect?

Which of the following metal chlorides has the highest covalent character?

Vedclass Test Series

Exam Paper Generator

Online Exam Module