A) \[\left( 0,0 \right)\]
B) \[\left( 1,1 \right)\]
C) \[\left( 1,-1 \right)\]
D) \[\left( -1,1 \right)\]
Correct Answer: B
Solution :
[b] Given curve is \[5{{x}^{2}}+{{y}^{2}}=1\] ......(i) To find the tangent, differentiate equation (i) \[5.2x+2y.\frac{dy}{dx}=0\] \[\Rightarrow \frac{dy}{dx}=\frac{-5x}{y}\Rightarrow {{\left( \frac{dy}{dx} \right)}_{at\left( \frac{1}{3},\frac{-2}{3} \right)}}=\frac{-5/3}{-2/3}=\frac{5}{2}\] So, equation of the tangent will be \[y-{{y}_{1}}=m(x-{{x}_{1}})\] \[\Rightarrow y+\frac{2}{3}=\frac{5}{2}\left( x-\frac{2}{3} \right)\Rightarrow 2(3y+2)=5(3x-1)\] \[\Rightarrow 15x-6y-5-4=0\text{ }\Rightarrow 15x-6y-9=0\] Since, point (1,1) satisfies the equation of the tangent of the given curve. Hence, option [b] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
