If P is the set of all real numbers alpha suc | vedclass

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

Question

(B) The given pair of straight lines is $3x^2+7xy+2y^2=0$. Factoring the equation,we get $(3x+y)(x+2y)=0$. The two lines are $L_1: 3x+y=0$ and $L_2: x

Vedclass · Accepted Answer

$-\frac{14}{3}$

Vedclass · Answer

(B) The given pair of straight lines is $3x^2+7xy+2y^2=0$. Factoring the equation,we get $(3x+y)(x+2y)=0$. The two lines are $L_1: 3x+y=0$ and $L_2: x+2y=0$. The lengths of perpendiculars from $(\alpha, 1)$ to $L_1$ and $L_2$ are $p_1 = \frac{|3\alpha+1|}{\sqrt{3^2+1^2}} = \frac{|3\alpha+1|}{\sqrt{10}}$ and $p_2 = \frac{|\alpha+2|}{\sqrt{1^2+2^2}} = \frac{|\alpha+2|}{\sqrt{5}}$. The product of the lengths is $p_1 p_2 = \frac{|(3\alpha+1)(\alpha+2)|}{\sqrt{50}} = \frac{|3\alpha^2+7\alpha+2|}{5\sqrt{2}}$. Given $p_1 p_2 = \frac{\sqrt{2}}{5}$,we have $\frac{|3\alpha^2+7\alpha+2|}{5\sqrt{2}} = \frac{\sqrt{2}}{5}$. This implies $|3\alpha^2+7\alpha+2| = 2$. Case $1$: $3\alpha^2+7\alpha+2 = 2 \implies 3\alpha^2+7\alpha = 0 \implies \alpha(3\alpha+7) = 0$. So $\alpha = 0$ or $\alpha = -\frac{7}{3}$. Case $2$: $3\alpha^2+7\alpha+2 = -2 \implies 3\alpha^2+7\alpha+4 = 0 \implies (3\alpha+4)(\alpha+1) = 0$. So $\alpha = -\frac{4}{3}$ or $\alpha = -1$. The set $P = \{0, -\frac{7}{3}, -\frac{4}{3}, -1\}$. The sum of elements is $0 - \frac{7}{3} - \frac{4}{3} - 1 = -\frac{11}{3} - 1 = -\frac{14}{3}$.

If $P$ is the set of all real numbers $\alpha$ such that the product of the lengths of perpendiculars from $(\alpha, 1)$ to the pair of straight lines $3x^2+7xy+2y^2=0$ is $\frac{\sqrt{2}}{5}$,then the sum of the elements in $P$ is

Similar Questions

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is:

If the equations of the opposite sides of a parallelogram are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$,then the equation of one of its diagonals is

If $(p, q)$ is the centroid of the triangle formed by the lines $8x^2 - 14xy + 5y^2 = 0$ and $x - 2y + 3 = 0$,then

If the combined equation of the diagonals of the square formed by the pairs of lines $xy+4x-5y-20=0$ and $xy-5x+4y-20=0$ is $x^2-y^2-kx+ly=0$,then $k+l=$.

If the pairs of straight lines represented by $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$ form a square,then $(h, c) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module