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Question: Find the point on the curve\[y = 3{x^2} + 4\] at which the tangent is perpendicular to the line whos...

Find the point on the curvey=3x2+4y = 3{x^2} + 4 at which the tangent is perpendicular to the line whose slope is −16   - \dfrac{1}{6}\;​.

Explanation

Solution

Let us find the slope of tangent of the given curve by differentiating the equation of given curve with respect to x. Then multiply the slope of the tangent with the slope of the given line and equate it with -1. Once we find the value x by using above correlation then we will put the value of x in the curve such that we will get the value of y. Just write the value of x and y in coordinate form to get the point.

Complete step-by-step solution:
Let (x,y) be the points.
Given,
The slope of the given line =−16 = - \dfrac{1}{6}
We know if two lines are perpendicular then their multiplication of slope is equal to -1
∴ Slope of the line perpendicular to the given line
=−1−16= \dfrac{{ - 1}}{{ - \dfrac{1}{6}}}
=6= 6 ………(i)
Now the equation of the curve given
⇒\Rightarrow y=3x2+4y = 3{x^2} + 4
Differentiate the above equation both side w.r.t x,
\Rightarrow$$$\dfrac{{dy}}{{dx}} = 3 \times 2x$$ \Rightarrow[\dfrac{{dy}}{{dx}} = 6x\] Slope of the tangent at (x,y) \Rightarrow\dfrac{{dy}}{{dx}} = 6x$$ ……….(ii) Equating the both slopes or equating the equation (i) and (ii) Slope of the tangent at (x,y) = Slope of a line perpendicular to a given line. $\Rightarrow6x = 6 $\Rightarrow$$$x = 1
Now, put the value of x in the given curve
⇒\Rightarrow$$y = 3{x^2} + 4 = 3{\left( 1 \right)^2} + 4 = 3 + 4 = 7$$
∴ The required point is (1,7).

Additional information: A tangent line is a line that touches a curve at a single point and does not cross through it. The point where the curve and the tangent meet is called the point of tangency. We know that for a line y=mx+cy = mx + c its slope at any point is m. The same applies to a curve. When we say the slope of a curve, we mean the slope of tangent to the curve at a point.
To find the slope m of a curve at a particular point, we differentiate the equation of the curve. If the given curve is y=f(x), we evaluate dydx\dfrac{{dy}}{{dx}} or f′(x)f'(x)and substitute the value of x to find the slope.

Note: The slope of a linear function is always constant while the slope of a curve is called slope of tangent to the curve and it varies point to point.