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10-2. Parabola, Ellipse, Hyperbola
easy
The equation of the tangent parallel to $y - x + 5 = 0$ drawn to $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is
A
$x - y - 1 = 0$
B
$x - y + 2 = 0$
C
$x + y - 1 = 0$
D
$x + y + 2 = 0$
Solution
(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 \times 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.
Standard 11
Mathematics
