Mathematics - Parabola Question with Solution | TestHub

Let the equation of the parabola be $y^2=4ax$. Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ be the extremities of the chord $PQ$. The coordinates of $T$, the point of intersection of the tangents at $P$ and $Q$, are $(a{t}_1{t}_2,a(t_1+t_2))$.

Now, $SP=a(1+t_1^2)$ and $SQ=a(1+t_2^2)$.

And $S{T}^2=(a{t}_1{t}_2-a{)}^2+(a(t_1+t_2)-0{)}^2$

$S{T}^2=a^2(t_1{t}_2-1{)}^2+a^2(t_1+t_2{)}^2$

$S{T}^2=a^2(t_1^2{t}_2^2-2t_1{t}_2+1+t_1^2+2t_1{t}_2+t_2^2)$

$S{T}^2=a^2(t_1^2{t}_2^2+t_1^2+t_2^2+1)$

$S{T}^2=a^2(1+t_1^2)(1+t_2^2)$

So, $S{T}^2=SP\cdot SQ$.

Therefore, $SP$, $ST$, and $SQ$ are in geometric progression (GP).