IIT-JEE 1994Question 1
Which one of the following curves cuts the parabola $y^{2}=4ax$ at right angles?
Differentiation and its Application
Solution
Concept. At a common point, compute both slopes and their product. Formulas used. (1) Parabola $y^2=4ax$: slope $\dfrac{2a}{y}$; right angle iff $m_1m_2=-1$Step 1 - (B). $y=e^{-x/2a}$ has slope $-\frac{y}{2a}$. Product with (1): $\frac{2a}{y}\cdot\left(-\frac{y}{2a}\right)=-1$ at every common point. Right angles. Step 2 - The others. (A) Circle slope $-\frac xy$; product $-\frac{2ax}{y^2}=-\frac12$. (C) $y=ax$ meets the parabola at $\left(\frac4a,4\right)$; product $\frac{2a}{4}\cdot a=\frac{a^2}{2}>0$. (D) $x^2=4ay$ meets it at $(4a,4a)$; slopes $\frac12$ and $2$, product $1$. Answer: (B)