The equation of the perpendiculars drawn from the origin to the lines represented by the equation $2 x^{2} - 10 x y + 12 y^{2} + 5 x - 16 y - 3 = 0$ is

$6 x^{2} + 5 x y + y^{2} = 0$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$6 y^{2} + 5 x y + x^{2} = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$6 x^{2} - 5 x y + y^{2} = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$6 x^{2} + 5 x y + y^{2} = 0$

Explanation for the correct option:
Step 1: Compare given equation for pair of straight lines with standard form
Comparing given equation for pair of straight lines with standard form we get,
$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ we get,
$a = 2, h = - 5, b = 12 . g = \frac{5}{2}, f = - 8, c = - 3$
Step 2: Use formula for sum of slopes
We know the sum of the slopes of the lines is given as
$m_{1} + m_{2} = \frac{- 2 h}{b}$
$\Rightarrow$ $m_{1} + m_{2} = \frac{10}{12}$
$\Rightarrow$ $m_{1} + m_{2} = \frac{5}{6}$ $. . . (i)$
Step 3: Use formula for product of slopes
We know the product of the slopes of the lines is given as
$m_{1} m_{2} = \frac{a}{b}$
$\Rightarrow$ $m_{1} m_{2} = \frac{2}{12}$
$\Rightarrow$ $m_{1} m_{2} = \frac{1}{6}$
$\Rightarrow$ $m_{1} = \frac{1}{6 m_{2}}$ $. . . (i i)$
Step 4: Solve for $m_{1}$ and $m_{2}$
Substitute value of $m_{1}$ from $(i i)$ in $(i)$
$\frac{1}{6 m_{2}} + m_{2} = \frac{5}{6}$
$6 {m_{2}}^{2} - 5 m_{2} + 1 = 0$
$\Rightarrow$ $6 {m_{2}}^{2} - 3 m_{2} - 2 m_{2} + 1 = 0$
$\Rightarrow$ $3 m_{2} (2 m_{2} - 1) - 1 (2 m_{2} - 1) = 0$
$\Rightarrow$ $(3 m_{2} - 1) (2 m_{2} - 1) = 0$
$\Rightarrow$ $m_{2} = \frac{1}{3} o r m_{2} = \frac{1}{2}$
$\Rightarrow$ $m_{1} = \frac{1}{2} o r m_{1} = \frac{1}{3}$
The slopes of the perpendicular lines are negative reciprocals
$∴$ The slopes of perpendicular lines are $- 2$ and $- 3$
Step 5: Write the equations of the lines passing through origin
Hence the equations of the lines are
$y = - 2 x$ (line passing through origin)
$\Rightarrow$ $y + 2 x = 0$ $. . . (i i i)$
$y = - 3 x$ (line passing through origin)
$\Rightarrow$ $y + 3 x = 0$ $. . . (i v)$
Step 6: Write the combined equations of the lines
The combined equation of these lines is given as
$(2 x + y) (3 x + y) = 0$
$\Rightarrow$ $6 x^{2} + 5 x y + y^{2} = 0$
Hence , option (A) is the correct answer.

Suggest Corrections

Similar questions

2

λ x^{2} - 10 x y + 12 y^{2} + 5 x - 16 y - 3 = 0

λ

k x^{2} - 10 x y + 12 y^{2} + 5 x - 16 y - 3 = 0

K

| λ | < 16

2 x^{2} - 10 x y + 12 y^{2} + 5 x + λ y - 3 = 0

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

2 x^{2} - 10 x y + 26 y^{2} + 5 x - 16 y - 3 = 0

Join BYJU'S Learning Program