A) \[y=-\frac{x}{m}\pm a\sqrt{1+{{m}^{2}}}\]
B) \[x+my=\pm \text{ }a\text{ }\sqrt{1+{{m}^{2}}}\]
C) \[x+my=\pm a\sqrt{1+{{(1/m)}^{2}}}\]
D) \[x-my=\pm a\sqrt{1+{{m}^{2}}}\]
Correct Answer: B
Solution :
Line perpendicular to \[y=mx+c\] is \[y=-\frac{1}{m}x+\lambda \] and \[m\lambda =\pm a\sqrt{1+{{m}^{2}}}\] Hence required tangent is \[my+x=\pm a\sqrt{1+{{m}^{2}}}\].You need to login to perform this action.
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