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 given equation is $ax^2+2hxy-2ay^2+3x+15y-9=0$. Since the lines intersect at $(1,1)$,the point $(1,1)$ must satisfy the equation: $a(1)^2+2h(1

Vedclass · Accepted Answer

-$7$

Vedclass · Answer

(C) The given equation is $ax^2+2hxy-2ay^2+3x+15y-9=0$. Since the lines intersect at $(1,1)$,the point $(1,1)$ must satisfy the equation: $a(1)^2+2h(1)(1)-2a(1)^2+3(1)+15(1)-9=0$ $a+2h-2a+3+15-9=0$ $-a+2h+9=0 \implies a-2h=9$ (Equation $1$). For a general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ to represent a pair of lines,the condition is $ABC+2FGH-AF^2-BG^2-CH^2=0$. Here,$A=a, H=h, B=-2a, G=3/2, F=15/2, C=-9$. Substituting these values: $a(-2a)(-9) + 2(15/2)(h)(3/2) - a(15/2)^2 - (-2a)(3/2)^2 - (-9)(h)^2 = 0$ $18a^2 + \frac{45h}{2} - \frac{225a}{4} + \frac{18a}{4} + 9h^2 = 0$ $18a^2 + 9h^2 + \frac{45h}{2} - \frac{207a}{4} = 0$ Using $a=2h+9$ from Equation $1$,substituting into the condition and solving for $h$ yields $h=-3$ and $a=3$. Thus,$ah = 3 	imes (-3) = -9$. Wait,re-evaluating the intersection point condition: the partial derivatives with respect to $x$ and $y$ must be zero at $(1,1)$. $f_x = 2ax+2hy+3 = 0 \implies 2a+2h+3=0$ $f_y = 2hx-4ay+15 = 0 \implies 2h-4a+15=0$ Solving these: $2a+2h=-3$ and $-4a+2h=-15$. Subtracting: $6a=12 \implies a=2$. Then $2(2)+2h=-3 \implies 2h=-7 \implies h=-3.5$. $ah = 2 	imes (-3.5) = -7$.

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

Similar Questions

If the bisectors of the angles represented by $ax^2 + 2hxy + by^2 = 0$ and $a'x^2 + 2h'xy + b'y^2 = 0$ are the same,then:

If $r(1 - m^2) + m(p - q) = 0$,then a bisector of the angle between the lines represented by the equation $px^2 - 2rxy + qy^2 = 0$ is

If $y = mx$ is one of the bisectors of the angle between the lines $ax^2 - 2hxy + by^2 = 0$,then

The square of the distance from the origin to the point of intersection of the pair of lines $ax^2+2hxy-ay^2+2gx+2fy+c=0$ is

The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2y=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module