The product of the lengths of the perpendicul | vedclass

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

Question

(A) The given equation of the pair of lines is $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$. Comparing this with the general form $ax^2 + by^2 + 2hxy + 2gx +

Vedclass · Accepted Answer

$\frac{3}{\sqrt{20}}$

Vedclass · Answer

(A) The given equation of the pair of lines is $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$. Comparing this with the general form $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$,we get $a = 1, b = 3, h = 2, g = -2, f = -5, c = 3$. The product of the lengths of the perpendiculars from the origin $(0, 0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the formula $\frac{|c|}{\sqrt{(a-b)^2 + (2h)^2}}$. Substituting the values,we get $\frac{|3|}{\sqrt{(1-3)^2 + (2 	imes 2)^2}} = \frac{3}{\sqrt{(-2)^2 + 4^2}} = \frac{3}{\sqrt{4 + 16}} = \frac{3}{\sqrt{20}}$.

The product of the lengths of the perpendiculars from the origin to the pair of lines $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$ is

Similar Questions

The value of $k$,if the slope of one of the lines given by $4x^2 + kxy + y^2 = 0$ is four times that of the other,is given by

All the points $(x, y)$ in the plane satisfying the equation $x^2+2x \sin(xy)+1=0$ lie on

The product of the perpendicular distances drawn from the origin to the pair of straight lines $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$ is

If two lines represented by $ax^2+2hxy+by^2=0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\tan(\alpha+\beta)=$

If the lines represented by the equation $4xy + 6x - 8y + c = 0$ form a rectangle with the coordinate axes,then the area of the rectangle (in sq. units) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module