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

$A$ bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill,when the bucket is at the highest position? (Take $g = 10\, m/sec^2$)

Where is the International Rice Research Institute $(IRRI)$ located?

$N$ divisions on the main scale of a vernier calipers coincide with $(N + 1)$ divisions of the vernier scale. If each division of main scale is $a$ units, then the least count of the calipers is

Which of the following forms a colloidal solution in water?

Persistent nucellus in the seed is known as

Vedclass Test Series

Exam Paper Generator

Online Exam Module