If the pair of lines 3x^2 - 5xy + py^2 = 0 an | vedclass

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

Question

(C) The second equation is $6x^2 - xy - 5y^2 = 0$. Factoring this,we get $(6x + 5y)(x - y) = 0$. So the lines are $y = -\frac{6}{5}x$ and $y = x$. If

Vedclass · Accepted Answer

$2, -\frac{25}{4}$

Vedclass · Answer

(C) The second equation is $6x^2 - xy - 5y^2 = 0$. Factoring this,we get $(6x + 5y)(x - y) = 0$. So the lines are $y = -\frac{6}{5}x$ and $y = x$. If $y = x$ is a common line,it must satisfy $3x^2 - 5x(x) + p(x)^2 = 0$,which gives $3x^2 - 5x^2 + px^2 = 0$,so $px^2 = 2x^2$,implying $p = 2$. If $y = -\frac{6}{5}x$ is a common line,it must satisfy $3x^2 - 5x(-\frac{6}{5}x) + p(-\frac{6}{5}x)^2 = 0$. This simplifies to $3x^2 + 6x^2 + p(\frac{36}{25})x^2 = 0$,which gives $9x^2 + \frac{36p}{25}x^2 = 0$. Dividing by $x^2$,we get $9 + \frac{36p}{25} = 0$,so $\frac{36p}{25} = -9$,which means $p = -9 	imes \frac{25}{36} = -\frac{25}{4}$. Thus,$p = 2$ or $p = -\frac{25}{4}$.

If the pair of lines $3x^2 - 5xy + py^2 = 0$ and $6x^2 - xy - 5y^2 = 0$ have one line in common,then $p =$

Similar Questions

The combined equation of two lines passing through the origin and making an angle of $45^{\circ}$ with the line $3x + y = 0$ is:

If $4ab = 3h^2$,then the ratio of slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

Find the length of the intercept cut by the pair of lines $2x^2 + 4xy - 4y^2 - 6x - 8y + 7 = 0$ on the $Y$-axis.

Two of the lines represented by the equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ will be perpendicular,then

If one of the lines given by $6x^2 - xy + 4cy^2 = 0$ is $3x + 4y = 0$,then $c$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module