Question

For which value of k, the equation
x^2-2xy+5y^2+kx-5y+16=0 will represent
pair of
of straight lines.
​

Answer 1

We're given the equation of a pair of straight lines,

$\longrightarrow x^2-2xy+5y^2+kx-5y+16=0$

$\longrightarrow5y^2-(2x+5)y+(x^2+kx+16)=0$

Equating discriminant to zero,

$\longrightarrow [-(2x+5)]^2-20(x^2+kx+16)=0$

$\longrightarrow 4x^2+20x+25-20x^2-20kx-320= 0$

$\longrightarrow -16x^2+20(1-k)x-295=0$

$\longrightarrow 16x^2+20(k-1)x+295=0$

Again equating discriminant of the polynomial in the LHS to zero,

$\longrightarrow [20(k-1)]^2-4\cdot16\cdot295=0$

$\longrightarrow 400k^2-800k+400-18880=0$

$\longrightarrow 5k^2-10k-231=0$

$\longrightarrow k=\dfrac{10\pm\sqrt{(-10)^2-4\cdot5\cdot-231}}{2\cdot5}$

$\longrightarrow k=\dfrac{10\pm\sqrt{100+4620}}{10}$

$\longrightarrow\underline{\underline{k=1\pm\sqrt{47.2}}}$

Answer 2

Answer:

We're given the equation of a pair of straight lines,

\longrightarrow x^2-2xy+5y^2+kx-5y+16=0⟶x

2

−2xy+5y

2

+kx−5y+16=0

\longrightarrow5y^2-(2x+5)y+(x^2+kx+16)=0⟶5y

2

−(2x+5)y+(x

2

+kx+16)=0

Equating discriminant to zero,

\longrightarrow [-(2x+5)]^2-20(x^2+kx+16)=0⟶[−(2x+5)]

2

−20(x

2

+kx+16)=0

\longrightarrow 4x^2+20x+25-20x^2-20kx-320= 0⟶4x

2

+20x+25−20x

2

−20kx−320=0

\longrightarrow -16x^2+20(1-k)x-295=0⟶−16x

2

+20(1−k)x−295=0

\longrightarrow 16x^2+20(k-1)x+295=0⟶16x

2

+20(k−1)x+295=0

Again equating discriminant of the polynomial in the LHS to zero,

\longrightarrow [20(k-1)]^2-4\cdot16\cdot295=0⟶[20(k−1)]

2

−4⋅16⋅295=0

\longrightarrow 400k^2-800k+400-18880=0⟶400k

2

−800k+400−18880=0

\longrightarrow 5k^2-10k-231=0⟶5k

2

−10k−231=0

\longrightarrow k=\dfrac{10\pm\sqrt{(-10)^2-4\cdot5\cdot-231}}{2\cdot5}⟶k=

2⋅5

10±

(−10)

2

−4⋅5⋅−231

\longrightarrow k=\dfrac{10\pm\sqrt{100+4620}}{10}⟶k=

10

10±

100+4620

\longrightarrow\underline{\underline{k=1\pm\sqrt{47.2}}}⟶

k=1±

47.2