JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
Let $A(0,a+2)$, $B(0,a)$, $C(-2,0)$ and $D(2,0)$ be four points and the lines $AD$ and $BC$ intersect at $P(x,y)$. If the locus of $P$ is the curve $f(x,y)=0$ and the tangent at the point $(4,\gamma)$ on this curve is $\frac x\alpha+\frac y\beta=1$, then $\alpha(\beta-\gamma)$ is equal to ________.
Points and Straight Lines
Answer$4$
Solution
$AD$: $\frac x2+\frac y{a+2}=1$, $BC$: $-\frac x2+\frac ya=1$. Solving: $x=\frac2{a+1}$, $y=\frac{a(a+2)}{a+1}=(a+1)-\frac1{a+1}=\frac2x-\frac x2$. So the locus is $x^2+2xy=4$. At $x=4$: $16+8\gamma=4\Rightarrow\gamma=-\frac32$. Tangent at $(4,\gamma)$: $4x+(\gamma x+4y)=4\Rightarrow\frac52x+4y=4\Rightarrow\frac{x}{8/5}+\frac y1=1$, so $\alpha=\frac85,\beta=1$. $\alpha(\beta-\gamma)=\frac85\cdot\frac52=4$.