If ax^2+2hxy-2ay^2+3x+15y-9=0 represents a pa | vedclass

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

Question

(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. Given the equation $ax^2+2hxy-2ay^2+3x+15y-9=0$,we have $b = -2a$,$2g = 3

Vedclass · Accepted Answer

-$7$

Vedclass · Answer

(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. Given the equation $ax^2+2hxy-2ay^2+3x+15y-9=0$,we have $b = -2a$,$2g = 3$,$2f = 15$,and $c = -9$.Since the lines intersect at $(x_0, y_0) = (1, 1)$,the partial derivatives with respect to $x$ and $y$ must be zero at this point.$\frac{\partial}{\partial x} (ax^2+2hxy-2ay^2+3x+15y-9) = 2ax+2hy+3$. At $(1, 1)$,$2a+2h+3 = 0 \implies 2a+2h = -3$.$\frac{\partial}{\partial y} (ax^2+2hxy-2ay^2+3x+15y-9) = 2hx-4ay+15$. At $(1, 1)$,$2h-4a+15 = 0 \implies 2h-4a = -15$.Subtracting the two equations: $(2a+2h) - (2h-4a) = -3 - (-15) \implies 6a = 12 \implies a = 2$.Substituting $a=2$ into $2a+2h = -3$: $2(2)+2h = -3 \implies 4+2h = -3 \implies 2h = -7 \implies h = -3.5$.Thus,$ah = 2 	imes (-3.5) = -7$.

If $ax^2+2hxy-2ay^2+3x+15y-9=0$ represents a pair of lines intersecting at $(1,1)$,then $ah=$

Similar Questions

The work functions of metals $A$ and $B$ are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of photoelectrons emitted is (given that $f > f_{0A}$ and $2f > f_{0B}$).

The major product of the following reaction is

Which of the following oxides cannot act as a reducing agent?

Which has the maximum number of molecules among the following?

An oxide of an element is a gas and dissolves in water to give an acidic solution. The element belongs to

Vedclass Test Series

Exam Paper Generator

Online Exam Module