The equation x^2-5xy+py^2+3x-8y+2=0 represent | vedclass

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

Question

(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5

Vedclass · Accepted Answer

$\frac{1}{\sqrt{50}}$

Vedclass · Answer

(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5}{2}, b=p, g=\frac{3}{2}, f=-4, c=2$.The equation represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.Substituting the values: $1(p)(2) + 2(-4)(\frac{3}{2})(-\frac{5}{2}) - 1(-4)^2 - p(\frac{3}{2})^2 - 2(-\frac{5}{2})^2 = 0$.$2p + 30 - 16 - \frac{9p}{4} - \frac{25}{2} = 0$.Multiplying by $4$: $8p + 120 - 64 - 9p - 50 = 0$ $\Rightarrow -p + 6 = 0$ $\Rightarrow p=6$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$.$	an 	heta = \left| \frac{2\sqrt{(-\frac{5}{2})^2 - 1(6)}}{1+6} ight| = \left| \frac{2\sqrt{\frac{25}{4}-6}}{7} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{7} ight| = \frac{2(\frac{1}{2})}{7} = \frac{1}{7}$.Since $	an 	heta = \frac{1}{7}$,we have a right triangle with opposite side $1$ and adjacent side $7$. The hypotenuse is $\sqrt{1^2+7^2} = \sqrt{50}$.Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{1}{\sqrt{50}}$.

The equation $x^2-5xy+py^2+3x-8y+2=0$ represents a pair of straight lines. If $\theta$ is the angle between them,then $\sin \theta$ is equal to

Similar Questions

The angle between the pair of straight lines $x^2 + 4y^2 - 7xy = 0$ is

If the angle between the lines represented by the equation $y^2 + kxy - x^2 \tan^2 A = 0$ is $2A$,then $k =$

The equation $4x^2 + 12xy + 9y^2 + 2gx + 2fy + c = 0$ represents two real parallel straight lines,if

The angle between the lines represented by the equation $y^{2} \sin^{2} \theta - xy \sin^{2} \theta + x^{2}(\cos^{2} \theta - 1) = 0$ is

If $\theta$ is the angle between the two lines represented by $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,then what is the value of $\text{cosec}^2\theta$? (where $\lambda$ is a real number.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module