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10-2. Parabola, Ellipse, Hyperbola
easy
The equation of the tangents to the conic $3{x^2} - {y^2} = 3$ perpendicular to the line $x + 3y = 2$ is
A
$y = 3x \pm \sqrt 6 $
B
$y = 6x \pm \sqrt 3 $
C
$y = x \pm \sqrt 6 $
D
$y = 3x \pm 6$
Solution
(a) Tangent to $\frac{{{x^2}}}{1} – \frac{{{y^2}}}{3} = 1$
and perpendicular to $x + 3y – 2 = 0$
is given by $y = 3x \pm \sqrt {9 – 3} = 3x \pm \sqrt 6 $.
Standard 11
Mathematics