JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^2-2x-2y=1$ meet at the point $P$. Then the area (in unit$^2$) of the triangle $PAB$ is:
Parabola
Solution
Concept. $AB$ is the chord of contact of $P$. Substitute $A$ and $B$ into it to find $P$. Formulas used. (1) Chord of contact of $(h,k)$ for $y^2-2x-2y=1$: $yk-(x+h)-(y+k)=1$.(2) Area $=\frac12\times$ base $\times$ height.Step 1 - Through A(1, 3). $3k-(1+h)-(3+k)=1\Rightarrow2k-h=5$. Step 2 - Through B(1, -1). $-k-(1+h)-(-1+k)=1\Rightarrow-2k-h=1$. Step 3 - Solve. Adding: $-2h=6$, $h=-3$; then $k=1$. $P(-3,1)$. Step 4 - Use (2). $AB$ lies on $x=1$, $AB=4$; height $=1-(-3)=4$. Area $=\frac12\cdot4\cdot4=8$.Answer: D