Q. 134.5( 2 Votes )
Find a point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slope is 
Answer :
Given:
The curve y = 3x2 + 4 and the Slope of the tangent is 
y = 3x2 + 4
Differentiating the above w.r.t x
⇒ 
= 2
3x2 – 1 + 0
⇒ 
= 6x ...(1)
Since, tangent is perpendicular to the line,
The Slope of the normal = 
i.e,
= 
⇒ 
= 
⇒ x = 1
Substituting x = 1 in y = 3x2 + 4,
⇒ y = 3(1)2 + 4
⇒ y = 3 + 4
⇒ y = 7
Thus, the required point is (1,7).
Rate this question :
Tangents & Normals (Concept Builder Class)55 mins
Test your knowledge of Tangents & Normals (Quiz)52 mins
Medelian disorder series: Thalassemia, Sickle cell anaemia & Phenylketonuria60 mins
Meselson and Stahl experiment49 mins
Interactive Quiz on Molecular basis of Inheritance-0234 mins
Interactive Quiz on Human Evolution33 mins
Rapid Quiz challenge - Cell Structure & Cell Division52 mins
Different high order thinking questions based on DNA structure39 mins
Everything About Genetic Disorders55 mins
Human Evolution | An Important Concept40 mins
Find the equation of the tangent to the curve 
at the point (a2/4, a2/4)
Find the equation of tangent to the curve 
, at the point, where it cuts the x-axis.
Find the equation of the tangent and the normal to the following curves at the indicated points:
y = 2x2 – 3x – 1 at (1, – 2)
RD Sharma - Volume 1Find the equation of the tangent and the normal to the following curves at the indicated points:
y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)
RD Sharma - Volume 1Find the equation of the tangent and the normal to the following curves at the indicated points:
y = x2 at (0, 0)
RD Sharma - Volume 1Find the equation of the tangent and the normal to the following curves at the indicated points:
y = x4 – 6x3 + 13x2 – 10x + 5 at x = 1 y = 3
RD Sharma - Volume 1Find the equation of the tangent and the normal to the following curves at the indicated points:
at (2, – 2)
Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally.
Mathematics - Exemplar