The equation of the tangent parallel to y - x | vedclass

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Question

(a) Given hyperbola is, $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$.....$(i)$

Equation of tangent parallel to $y - x + 5 = 0$ is

$y - x + \lambd

Vedclass · Accepted Answer

$x - y - 1 = 0$

Vedclass · Answer

(a) Given hyperbola is, $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$.....$(i)$

Equation of tangent parallel to $y - x + 5 = 0$ is

$y - x + \lambda = 0$

$ \Rightarrow $ $y = x - \lambda $.....$(ii)$

If line $(ii)$ is a tangent to hyperbola $(i),$ then

$ - \lambda = \pm \sqrt {3 	imes 1 - 2} $   (from $c = \pm \sqrt {{a^2}{m^2} - {b^2}} $)

$ - \lambda = \pm \,1$

$\Rightarrow \lambda = - 1,\, + 1$.

Put the values of $\lambda $ in $(ii),$

we get $x - y - 1 = 0$ and $x - y + 1 = 0$ are the required tangents.

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is

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