(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using

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(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using $(i)$:$3 x^2-x y+3 y^2+(2 x-3 y)\left(\frac{2 x+3 y}{k}\right)-4\left(\frac{2 x+3 y}{k}\right)^2=0$Multiplying by $k^2$:$k^2(3 x^2-x y+3 y^2) + k(4 x^2+6 x y-6 x y-9 y^2) - 4(4 x^2+9 y^2+12 x y) = 0$$x^2(3 k^2+4 k-16) + x y(-k^2-48) + y^2(3 k^2-9 k-36) = 0$Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(3 k^2+4 k-16) + (3 k^2-9 k-36) = 0$$6 k^2-5 k-52 = 0$

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

AP EAMCET 2023 All AP EAMCET

A
$6 k^2+5 k+52=0$
B
$6 k^2+5 k-52=0$
C
$6 k^2-5 k+52=0$
D
$6 k^2-5 k-52=0$

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Pair of straight lines

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Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

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The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k (>1)$ is

DifficultTS EAMCET 2015

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The line $3x + 4y - 5 = 0$ cuts the curve $2x^2 + 3y^2 = 5$ at $A$ and $B$. If $O$ is the origin,then $\angle AOB =$

EasyAP EAMCET 2019

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The lines joining the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$ to the origin are perpendicular. Then the value of $\lambda$ is:

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If the lines joining the origin to the points of intersection of the line $fx - gy = \lambda$ and the curve $x^2 + hxy - y^2 + gx + fy = 0$ are mutually perpendicular,then:

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If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

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If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

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