In the XY-plane- three distinct lines l1- l2- l3 concur at a point - 0- Further- the lines l1- l2- l3 are normals to the parabola y2 - 6x at points A - x1- y1- B - x2- y2- C - x3- y3- respectively- Then-

The correct option is B λ > 3 General Equation of normal of y ^ 2 =4ax #10;is y=mx 2am a m ^ 3 #10; #10;For y ^ 2 =6x , a= 3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Directrix of Hyperbola

In the XY-pla...

Question

In the XY-plane, three distinct lines $l_{1}, l_{2}, l_{3}$ concur at a point $(λ, 0)$ . Further, the lines $l_{1}, l_{2}, l_{3}$ are normals to the parabola $y^{2} = 6 x$ at points $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ , respectively. Then,

λ < - 5

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

λ > 3

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

- 5 < λ < - 3

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

0 < λ < 3

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

Open in App

Solution

The correct option is B $λ > 3$
General Equation of normal of $y^{2} = 4 a x$ is $y = m x - 2 a m - a m^{3}$

For $y^{2} = 6 x$ , $a = \frac{3}{2}$

Normal passes through $(λ, 0)$

$∴ 0 = m λ - 2 a m - a m^{3}$

$m (λ - 2 a - a m^{2}) = 0 m = 0, λ - 2 a - a m^{2} = 0 λ - 2 a - a m^{2} = 0 \Rightarrow m^{2} = \frac{λ - 2 a}{a} m^{2} > 0 \Rightarrow \frac{λ - 2 a}{a} > 0 \frac{λ}{a} > 2 λ > 2 a λ > 2 \times \frac{3}{2} λ > 3$

Hence, option B is correct.

Suggest Corrections

Similar questions

Q. In the xy-plane, three distinct lines

l_{1}, l_{2}, l_{3}

are concurrent at

M (λ, 0)

. Also the lines

l_{1}, l_{2}, l_{3}

are normals to the parabola

y^{2} = 6 x

at the points

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

respectively. Then

Q. If the lines

L_{1} : λ^{2} x - y - 1 = 0

L_{2} : x - λ^{2} y + 1 = 0

L_{3} : x + y - λ^{2} = 0

pass through the same point the value(s) of

λ

equals

Q. Let

L_{1}

, and

L_{2}

denotes the lines

\to r =^i + λ (-^i + 2^j + 2^k), λ \in R

and

\to r = μ (2^i -^j + 2^k), μ \in R

respectively.
If

L_{3}

is a line which is perpendicular to both

L_{1}

and

L_{2}

and cuts both of them, then which of the following option describe(s)

L_{3}

Q. AB, AC are tangents to a parabola

y^{2} = 4 a x

l_{1}, l_{2} . l_{3}

are the lengths of perpendiculars from A, B, C on any tangent to the parabola, then

Q. Three lines

L_{1} : \to r = λ^i, λ \in R,

L_{2} : \to r =^k + μ^j, μ \in R,

and

L_{3} : \to r =^i +^j + ν^k, ν \in R

are given.
For which point(s)

Q

L_{2}

can we find a point

P

L_{1}

and a point

R

L_{3}

so that

P, Q

and

R

are collinear ?

Join BYJU'S Learning Program

In the XY-plane, three distinct lines l1,l2,l3 concur at a point (λ,0). Further, the lines l1,l2,l3 are normals to the parabola y2=6x at points A=(x1,y1),B=(x2,y2),C=(x3,y3), respectively. Then,

The correct option is B λ>3General Equation of normal of y2=4ax is y=mx−2am−am3 For y2=6x , a=32 Normal passes through (λ,0) ∴0=mλ−2am−am3 m(λ−2a−am2)=0m=0,λ−2a−am2=0λ−2a−am2=0⇒m2=λ−2aam2>0⇒λ−2aa>0λa>2λ>2aλ>2×32λ>3 Hence, option B is correct.

In the XY-plane, three distinct lines $l_{1}, l_{2}, l_{3}$ concur at a point $(λ, 0)$ . Further, the lines $l_{1}, l_{2}, l_{3}$ are normals to the parabola $y^{2} = 6 x$ at points $A = (x_{1}, y_{1}), B = (x_{2}, y_{2}), C = (x_{3}, y_{3})$ , respectively. Then,