Question

If a triangle is formed by any three tangents of the parabola $y^2=4ax$ whose two of its vertices lie on $x^2=4by$, then third vertex lie on

• A. $(x-1{)}^2=4ay$
• B. $x^2=16ay$
• C. $x^2=4by$
• D. $(x+1{)}^2=4ay$

Answer 1

The correct option is C $x^2=4by$

Let the three tangent equations to parabola $y^2=4ax$ are
$y{t}_1=x+a{t}_1^2$
$y{t}_2=x+a{t}_2^2$
$y{t}_3=x+a{t}_3^2$
Intersection points of these tangents are $A(a{t}_1{t}_2,a(t_1+t_2\left)\right),B(a{t}_2{t}_3,a(t_2+t_3\left)\right),C(a{t}_1{t}_3,a(t_1+t_3\left)\right)$

Let $A$ and $B$ lie on $x^2=4by$

$a^2{t}_1^2{t}_2^2=4ab(t_1+t_2)\cdots \left(1\right)$
$a^2{t}_2^2{t}_3^2=4ab(t_2+t_3)\cdots \left(2\right)$
Now eliminating $t_3$ from above equations and hence bringing relationship between $t_1,t_2$

From $\left(1\right)÷\left(2\right)$, we get
$\frac{t_1^2}{t_3^2}=\frac{t_1+t_2}{t_2+t_3}$
$\Rightarrow t_2(t_1^2-t_3^2)=t_1{t}_3(t_3-t_1)$
$\Rightarrow t_2=-\frac{t_1{t}_3}{(t_1+t_3)}\cdots \left(3\right)$

From $\left(1\right)-\left(2\right)$

$a^2{t}_2^2(t_1^2-t_3^2)=4ab(t_1-t_3)$
$a^2{[-\frac{t_1{t}_3}{(t_1+t_3)}]}^2(t_1+t_3)=4ab\Rightarrow (a{t}_1{t}_3{)}^2=4b[a(t_1+t_3)]$
Hence $C(a{t}_1{t}_3,a(t_1+t_3\left)\right)$ will also lie on $x^2=4by$