If the lines x^2+2xy-35y^2-4x+44y-12=0 and 5x | vedclass

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

Question

(C) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$. The point of intersection $(x, y)$ of these lines is given by the formula: $x = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$ $y = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$ Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the equation $5x+ky-8=0$. Substituting the values: $5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$ $\frac{20}{3} + \frac{2k}{3} = 8$ $20 + 2k = 24$ $2k = 4$ $k = 2$

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+ky-8=0$ are concurrent,then $k$ equals

Similar Questions

Among the following compounds,which one is not responsible for the depletion of the ozone layer?

The curve satisfying the differential equation,$ydx - (x + 3y^2) dy = 0$ and passing through the point $(1, 1)$,also passes through the point

If $g$ is the acceleration due to gravity on the earth's surface,the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is

The total work done on a particle is equal to the change in its kinetic energy. This is applicable:

The extreme values of $4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$ over $\mathbb{R}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module