Equation of a tangent to hyperbola 4x^2 -5y^2 = 20 parallel to line x -y = 2 is

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10-2. Parabola, Ellipse, Hyperbola

hard

The equation of a tangent to the hyperbola $4x^2 -5y^2 = 20$ parallel to the line $x -y = 2$ is

A

$x -y + 1 = 0$

B

$x -y + 7 = 0$

C

$x -y + 9 = 0$

D

$x -y -3 = 0$

(JEE MAIN-2019)

Solution

Hyperbola is $\frac{{{x^2}}}{5} – \frac{{{y^2}}}{4} = 1$

Equation of its tangent in slop from is $y = mx \pm \sqrt {5{m^2} – 4} $

Hence tangent with slope $1$ is $y = x \pm 1$

Standard 11

Mathematics

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