Gujarati
Hindi
10-2. Parabola, Ellipse, Hyperbola
normal

A common tangent to $9x^2 + 16y^2 = 144 ; y^2 - x + 4 = 0 \,\,\&\,\, x^2 + y^2 - 12x + 32 = 0$ is :

A

$y = 3$

B

$x = - 4$

C

$x = 4$

D

$y = - 3$

Solution

$y = (1/2) x + 2\ \Rightarrow 4 = 4. 1/4 + b^2 $

$\Rightarrow b^2 = 3$ again $4 = 4m^2 + 3$
$ \Rightarrow m = ± 1/2 ; $   make a figure and interpret the result

Standard 11
Mathematics

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