Point of contact of line y = x - 1 with 3{x^2} - 4{y^2} = 12 is

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10-2. Parabola, Ellipse, Hyperbola

easy

The point of contact of the line $y = x - 1$ with $3{x^2} - 4{y^2} = 12$ is

A

$(4, 3)$

B

$(3, 4)$

C

$(4, -3)$

D

None of these

Solution

(a) The equation of the line and hyperbola are

$y = x – 1$…..$(i)$

$3{x^2} – 4{y^2} = 12$…..$(ii)$

From $(i)$ and $(ii),$ we get

$3{x^2} – 4{(x – 1)^2} = 12$

$ \Rightarrow 3{x^2} – 4({x^2} – 2x + 1) = 12$

or ${x^2} – 8x + 16 = 0$

$⇒$ $x = 4$

From $(i),$ $y = 3.$

So point of contact is $(4,3)$.

Standard 11

Mathematics

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