If t1 and t2 are the roots of the equation t2- t-1-0- where - is a parameter- then the line joining the points at12-2at1 and at22-2at2 always passes through

Since t_1 and t_2 are the roots of the given equation, ∴ t_1+t_2= λ ⋯(1) and t_1t_2=1 ⋯(2) Equation of the line joining the given points is nbsp; y 2a ...

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Byju's Answer

Standard XIII

Mathematics

Two Point Form of a Line

If t1 and t2 ...

Question

If $t_{1}$ and $t_{2}$ are the roots of the equation $t^{2} + λ t + 1 = 0$ , where $λ$ is a parameter, then the line joining the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ always passes through

(a, 0)

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(- a, 0)

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(λ, 0)

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(- λ, 0)

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Solution

The correct option is B $(- a, 0)$
Since $t_{1}$ and $t_{2}$ are the roots of the given equation,
$∴ t_{1} + t_{2} = - λ \dots (1)$
and $t_{1} t_{2} = 1 \dots (2)$
Equation of the line joining the given points is
$y - 2 a t_{1} = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} (x - a t_{1}^{2}) \Rightarrow y (t_{1} + t_{2}) = 2 x + 2 a t_{1} t_{2}$

Using equation $(1)$ and $(2)$ ,
$- λ y = 2 x + 2 a \Rightarrow y = (\frac{- 2}{λ}) (x + a)$
The above line always passes through $(- a, 0)$ .

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If t1 and t2 are the roots of the equation t2+λt+1=0, where λ is a parameter, then the line joining the points (at21,2at1) and (at22,2at2) always passes through

If $t_{1}$ and $t_{2}$ are the roots of the equation $t^{2} + λ t + 1 = 0$ , where $λ$ is a parameter, then the line joining the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ always passes through