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 given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = 5 \Rightarrow h = \frac{5}{2}$,and $b = 3$. The angle $	heta$ between the pair of straight 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| = \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 is $	an^{-1}(k)$,we have $	an 	heta = k$. Thus,$k = \pm \frac{1}{5}$.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

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 angle between the lines represented by the equation $ax^2 + xy + by^2 = 0$ will be $45^\circ$,if

If $\theta$ is the acute angle between the pair of lines $H \equiv ax^2 - xy + by^2 = 0$,$\tan \theta = 5$ and $(1, -1)$ is a point on $H = 0$,then $a^2 + ab + b^2 =$

The lines represented by the equation $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$ are

Which of the following equations represents a pair of straight lines perpendicular to each other?

Vedclass Test Series

Exam Paper Generator

Online Exam Module