Question

The point of intersection of the tangents at $t_1=t$ and $t_2=3t$ to the parabola $y^2=8x$ is:

• A. $(t^2,4t)$
• B. $(6t^2,8t)$
• C. $(4t,t^2)$
• D. $(8t,6t^2)$

Answer 1

Answer: (B)

$(6t^2,8t)$

Given equation:$y^2=8x$ ...$\left(1\right)$

Comparing the above equation with the general equation of parabola $y^2=4ax$

So,$a=2$

The parametric coordinates of any point on $\left(1\right)$ are $(a{t}^2,2at)$ or $(2t²,4t)$

Coordinates of the point $P$ are $P(2t_1^2.4t_1)$ and $Q$ are $Q(2t_2^2.4t_2)$.

Differentiate equation $\left(1\right)$ w.r.t $x$ we get,

$2y\frac{dy}{dx}=8\frac{dy}{dx}=\frac{4}{y}$

Slope of the tangent at point $P=\frac{4}{4t_1}=\frac{1}{t_1}$

Slope of the tangent at point $Q=\frac{4}{4t_2}=\frac{1}{t_2}$

Now, the equation of $RP$ is,

$\frac{(y-4t_1)}{(x-2t_1^2)}=\frac{1}{t_1}$

$\Rightarrow t_{1}y-4t_1^2=x-2t_1^2{t}_{1}y-x=2t_1^2$ ...$\left(2\right)$

Now, the equation of $RQ$ is,

$t_{2}y-x=2t_2^2$ ...$\left(2\right)$

Solve equation $\left(1\right)$ and $\left(2\right)$ we get

$x=2t_1{t}_2=2\times t\times 3t=6t^2$

$y=2(t_1+t_2)=2(t+3t)=8t$

$\therefore (x,y)=(6t^2,8t)$