JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
Let a conic $C$ pass through the point $(4,-2)$ and $P(x,y)$, $x\ge3$, be any point on $C$. Let the slope of the line touching the conic $C$ only at a single point $P$ be half the slope of the line joining the points $P$ and $(3,-5)$. If the focal distance of the point $(7,1)$ on $C$ is $d$, then $12d$ equals $\_\_\_\_\_\_$ .
Parabola
Answer$75$
Solution
Concept. The slope condition is a separable differential equation whose solution is a parabola with vertex $(3,-5)$. Formulas used. (1) $\displaystyle\int\frac{dy}{y+5}=\ln|y+5|$.(2) $(y-k)^2=4a(x-h)$: focal distance of a point $=(x-h)+a$.Step 1 - Equation. $\dfrac{dy}{dx}=\dfrac12\cdot\dfrac{y+5}{x-3}\Rightarrow\dfrac{2\,dy}{y+5}=\dfrac{dx}{x-3}$. Step 2 - Use (1). $2\ln|y+5|=\ln|x-3|+C\Rightarrow(y+5)^2=K(x-3)$. Step 3 - Through (4, -2). $9=K\cdot1\Rightarrow K=9$. So $(y+5)^2=9(x-3)$, $4a=9$, $a=\frac94$. Step 4 - Check (7, 1). $36=9\cdot4$. Yes. Step 5 - Use (2). $d=(7-3)+\frac94=\frac{25}{4}$, so $12d=75$. Answer: 75