JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
If the tangent to the curve, $y=e^x$ at a point $(c,e^c)$ and the normal to the parabola, $y^2=4x$ at the point $(1,2)$ intersect at the same point on the $x$-axis, then the value of $c$ is ________.
Differentiation and its Application
Answer$4$
Solution
Concept. Find where the normal meets the $x$-axis; the tangent to $y=e^x$ must meet it at the same point. Formulas used. (1) For $y^2=4x$: $y'=\dfrac2y$(2) The tangent to $y=e^x$ at $x=c$ meets the $x$-axis at $x=c-1$Step 1 - Normal. By (1), the tangent slope at $(1,2)$ is $1$, so the normal is $y-2=-(x-1)$. It meets the $x$-axis at $(3,0)$. Step 2 - Tangent. $y-e^c=e^c(x-c)$; with $y=0$, $x=c-1$ (this is (2)). Step 3 - Solve. $c-1=3$, so $c=4$. Answer: 4