Question:
The equation of the normal to the curve
\[
y=\log_e x
\]
at the point $P(1,0)$ is:
The equation of the normal to the curve
\[
y=\log_e x
\]
at the point $P(1,0)$ is:
Show Hint
The tangent and normal at a point are always perpendicular. Therefore:
\[
m_t\cdot m_n=-1
\]
Use this relation immediately after finding the derivative.
Updated On: May 29, 2026
- $2x+y=2$
- $x-2y=1$
- $x-y=1$
- $x+y=1$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
The slope of the tangent to a curve is obtained using differentiation:
\[
m_t=\frac{dy}{dx}
\]
The slope of the normal is the negative reciprocal of the tangent slope:
\[
m_n=-\frac{1}{m_t}
\]
Step 1: Finding the slope of the tangent.
Given curve: \[ y=\log_e x \] Differentiate with respect to $x$: \[ \frac{dy}{dx}=\frac1x \] At the point $(1,0)$: \[ m_t=\left.\frac1x\right|_{x=1}=1 \] Thus, the slope of the tangent is \[ m_t=1 \]
Step 2: Finding the slope of the normal.
The slope of the normal is: \[ m_n=-\frac1{m_t} \] \[ m_n=-1 \]
Step 3: Forming the equation of the normal.
Using point-slope form: \[ y-y_1=m(x-x_1) \] Substitute \[ (x_1,y_1)=(1,0), \qquad m=-1 \] \[ y-0=-1(x-1) \] \[ y=-x+1 \] Rearranging: \[ x+y=1 \] Hence, \[ \boxed{x+y=1} \]
Step 1: Finding the slope of the tangent.
Given curve: \[ y=\log_e x \] Differentiate with respect to $x$: \[ \frac{dy}{dx}=\frac1x \] At the point $(1,0)$: \[ m_t=\left.\frac1x\right|_{x=1}=1 \] Thus, the slope of the tangent is \[ m_t=1 \]
Step 2: Finding the slope of the normal.
The slope of the normal is: \[ m_n=-\frac1{m_t} \] \[ m_n=-1 \]
Step 3: Forming the equation of the normal.
Using point-slope form: \[ y-y_1=m(x-x_1) \] Substitute \[ (x_1,y_1)=(1,0), \qquad m=-1 \] \[ y-0=-1(x-1) \] \[ y=-x+1 \] Rearranging: \[ x+y=1 \] Hence, \[ \boxed{x+y=1} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Tangents and Normals Questions
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
- The value of \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \) is equal to:
- For the curve \( y = 3x^3 - 3x^2 + 1 \) at \( x = 1 \), find the equation of the tangent.
- Let the curve be represented by $x=2(\cos t+t\sin t)$, $y=2(\sin t-t\cos t)$. Then normal at any point '$t$' of the curve is at a distance of _____ units from the origin.
- If $y = 4x - 5$ is a tangent to the curve $y^{2} = px^{3} + q$ at $(2, 3)$, then $p-q$ is
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions