Question
Mathematics Question on Straight lines
Find the equation of the normal to the curve y=3x2+1, which passes through (2,13).
A
(A) x+12y+158=0
B
(B) x−12y−156=0
C
(C) 12x+y−156=0
D
(D) x+12y−158=0
Answer
(D) x+12y−158=0
Explanation
Solution
Explanation:
The slope of the tangent to the curve =dydxThe slope of normal to the curve =−1(dydx)Point-slope is the general form: y−y1=m(x−x1), Where m= slopeHere, y=3x2+1dydx=6xdydx|x=2=12Slope of normal to the curve =−1(dydx)=−112Equation of normal to curve passing through (2,13) is:y−13=−112(x−2)⇒12y−156=−x+2⇒x+12y−158=0Hence, the correct option is (D).