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Question: The equation of normal to the curve \(y=\tan x\)at the point \(\left( 0,0 \right)\)is- A) \(x+y=0\...

The equation of normal to the curve y=tan⁡xy=\tan xat the point (0,0)\left( 0,0 \right)is-
A) x+y=0x+y=0
B) x−y=0x-y=0
C) x+2y=0x+2y=0
D) None of these

Explanation

Solution

Here we will find the equation of normal to the given curve. For that, we will find the slope of tangent of the given curve at the given point. For the slope of normal, we will take the negative of the inverse of the slope of tangent. Then we will find the equation of normal using the given point and the slope of the normal. The equation which we will get will be the required equation of normal.

Complete step by step solution:
The given curve is y=tan⁡xy=\tan xand the given point is (0,0)\left( 0,0 \right)
We will first find the slope of the tangent at point(0,0)\left( 0,0 \right).
We will differentiate both side of the given equation with respect to x.
dydx=dtan⁡xdx\dfrac{dy}{dx}=\dfrac{d\tan x}{dx}
Here, dydx\dfrac{dy}{dx}means the slope of the tangent of the given curve.
We know the derivative of tan⁡x\tan x with respect to x is sec⁡2x{{\sec }^{2}}x
Therefore,
dydx=sec⁡2x\dfrac{dy}{dx}={{\sec }^{2}}x
We have to find the slope at point (0,0)\left( 0,0 \right), we will put the value of x here.
dydx∣(0,0)=sec⁡20=1{{\left. \dfrac{dy}{dx} \right|}_{(0,0)}}={{\sec }^{2}}0=1
Hence the slope of tangent at point (0,0)\left( 0,0 \right)is 1.
Now, we will find the slope of normal at the same point(0,0)\left( 0,0 \right) which is:-
Slope of normal =−1dydx∣(0,0)=−11=−1=-{{\dfrac{1}{\left. \dfrac{dy}{dx} \right|}}_{\left( 0,0 \right)}}=-\dfrac{1}{1}=-1
Hence, the slope of the normal at point (0,0)\left( 0,0 \right)is -1.
Now, we will find the equation of the normal at point(0,0)\left( 0,0 \right).
y−0=slope(x−0)y-0=slope\left( x-0 \right)
We will put the value of slope normal here.
y−0=−1(x−0)y-0=-1\left( x-0 \right)
On further simplification, we get
x+y=0x+y=0
Therefore, the required equation of the normal is x+y=0x+y=0

Hence, the correct option is A.

Note:
Since we have calculated the slope of the normal and tangent of a curve at the given point. So we need to know the meaning of these terms.

  1. A tangent of a curve is defined as a line that touches the given curve at only one point.
  2. A normal to a curve is the line which is perpendicular to the tangent which touches the curve at that point.
  3. Slope of the normal to a curve at a point is equal to the negative of the inverse of the slope of the tangent to a curve at the same point.