If the straight lines joining the origin and the points of intersection of the curve 5 x2+12 x y-6 y2+4 x-2 y+3=0 and x+k y-1=0 are equally inclined to the co-ordinate axis, then the value of k -

Question

If the straight lines joining the origin and the points of intersection of the curve 5 x2+12 x y-6 y2+4 x-2 y+3=0 and x+k y-1=0 are equally inclined to the co-ordinate axis, then the value of k - (A) is equal to 1 (B) is equal to -1 (C) is equal to 2 (D) does not exist in the set of real numbers

Tardigrade Admin · Accepted Answer

Homogenizing the curve with the help of the straight line.

5 x^{2} + 12 x y - 6 y^{2} + 4 x (x + k y) - 2 y (x + k y) + 3 (x + k y)^{2} = 0

12 x^{2} + (10 + 4 k + 6 k) x y + (3 k^{2} - 2 k - 6) y^{2} = 0

Lines are equally inclined to the coordinate axes

∴ coefficient of x y = 0

\Rightarrow 10 k + 10 = 0 \Rightarrow k = - 1

Q. If the straight lines joining the origin and the points of intersection of the curve $5 x^{2} + 12 x y - 6 y^{2} + 4 x - 2 y + 3 = 0$ and $x + k y - 1 = 0$ are equally inclined to the co-ordinate axis, then the value of $k$ -

is equal to 1

is equal to $- 1$

is equal to 2

does not exist in the set of real numbers

Q. If the straight lines joining the origin and the points of intersection of the curve 5x2+12xy−6y2+4x−2y+3=0 and x+ky−1=0 are equally inclined to the co-ordinate axis, then the value of k -

is equal to 1

is equal to −1

is equal to 2

does not exist in the set of real numbers

Homogenizing the curve with the help of the straight line. 5x2+12xy−6y2+4x(x+ky)−2y(x+ky)+3(x+ky)2=0 12x2+(10+4k+6k)xy+(3k2−2k−6)y2=0 Lines are equally inclined to the coordinate axes ∴ coefficient of xy=0 ⇒10k+10=0⇒k=−1

Q. If the straight lines joining the origin and the points of intersection of the curve $5 x^{2} + 12 x y - 6 y^{2} + 4 x - 2 y + 3 = 0$ and $x + k y - 1 = 0$ are equally inclined to the co-ordinate axis, then the value of $k$ -

is equal to $- 1$