A) \[3x-4y-19=0,\,\,3x-4y+31=0\]
B) \[4x+3y-19=0,\,\,4x+3y+31=0\]
C) \[4x+3y+19=0,\,\,4x+3y-31=0\]
D) \[3x-4y+19=0,3x-4y+31=0\]
Correct Answer: C
Solution :
Let equation of tangent be\[4x+3y+k=0\], then \[\sqrt{9+4+12}=\left| \frac{4(3)+3(-2)+k}{\sqrt{16+9}} \right|\] \[\Rightarrow 6+k=\pm 25\Rightarrow k=19\]and\[-31\]. Hence the tangents are \[4x+3y+19=0\] and \[4x+3y-31=0\].You need to login to perform this action.
You will be redirected in
3 sec