Gujarati
10-2. Parabola, Ellipse, Hyperbola
easy

If $y = mx + c$ is tangent on the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$, then the value of $c$ is

A

$0$

B

$3/m$

C

$ \pm \sqrt {9{m^2} + 4} $

D

$ \pm 3\sqrt {1 + {m^2}} $

Solution

(c) Here, $a = 3,\,\,b = 2$. 

By formula, ${c^2} = {b^2} + {a^2}{m^2}$

${c^2} = 4 + 9{m^2}$;  $c = \pm \sqrt {9{m^2} + 4} $.

Standard 11
Mathematics

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