The equation x^2-3xy+lambda y^2+3x-5y+2=0,whe | vedclass

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

Question

(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - c

Vedclass · Accepted Answer

$\sqrt{10}$

Vedclass · Answer

(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Comparing with $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,we have $a=1, h=-3/2, b=\lambda, g=3/2, f=-5/2, c=2$. Substituting these values: $(1)(\lambda)(2) + 2(-5/2)(3/2)(-3/2) - (1)(-5/2)^2 - (\lambda)(3/2)^2 - (2)(-3/2)^2 = 0$. $2\lambda + 45/4 - 25/4 - 9\lambda/4 - 9/2 = 0$. $2\lambda - 9\lambda/4 + 5 - 4.5 = 0 \implies -\lambda/4 + 0.5 = 0 \implies \lambda = 2$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Here $a=1, b=2, h=-3/2$. $	an 	heta = \left| \frac{2\sqrt{9/4 - 2}}{1+2} ight| = \frac{2\sqrt{1/4}}{3} = \frac{1}{3}$. Since $	an 	heta = 1/3$,then $\cot 	heta = 3$. $\operatorname{cosec}^2 	heta = 1 + \cot^2 	heta = 1 + 3^2 = 10$. Thus,$\frac{\operatorname{cosec}^2 	heta}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}$.

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$,where $\lambda$ is a real number,represents a pair of lines. If $\theta$ is the acute angle between the lines,then $\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}} = $

Similar Questions

If $\operatorname{Tan}^{-1}(2 \sqrt{10})$ is the angle between the lines $ax^2 + 4xy - 2y^2 = 0$ and $a \in \mathbb{Z}$,then the product of the slopes of the given lines is

The number of real values of $\alpha$ for which the pair of lines represented by $(\alpha^2+12|\alpha|) x^2+6 x y+(18-21|\alpha|) y^2=0$ are at right angles to each other,is

If the angle between the two lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1} m$,then $m = $

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

If the pair of straight lines $6x^2 - 5xy + y^2 = 0$ makes angles $\alpha$ and $\beta$ with the $X$-axis,then $\tan(\alpha - \beta) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module