Question

The point on the curve given by locus in part (2) which is at a minimum distance from the line $y^2=x-2$

• A. $(6,2)$
• B. $(\frac{9}{4},\frac{1}{2})$
• C. $(\frac{13}{2},\frac{3}{2})$
• D. $(2,0)$

Answer 1

Answer: (A)

$(6,2)$

The given equation of the parabola $y^2=8x\to \left(1\right)$

The above equation is of the form $y^2=4ax$

$\Rightarrow 4ax=8x$

$a=2$

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

Now, coordinates of the point P are $P(2t_1^2,4t_1)$ and $Q$ are $P(2t_2^2,4t_2)$

Now, differentiating $\left(1\right)$

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

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

Now, 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 equation of RP is $\frac{y-4t_1}{x-2t_1^2}=\frac{1}{t_1}$

$\Rightarrow y-1t-4t_1^2=x-2t-1^2$

$\Rightarrow t_{1}y-x=2t_1^2\to \left(2\right)$

now, equation of RQ is $t_{2}y-x=2t_2^2\to \left(3\right)$

solving $\left(2\right)$ and $\left(3\right)$

$\Rightarrow x=2t_1{t}_2$ $y=2(t_1+t_2)$

now, we have $t^2+at+2=0$

now $t_1$ and $t_2$ are roots of above equation -

$\Rightarrow t_1+t_2=-a$

$t_1=2=2$

so, $x=4$ $y=-2a$

coordinating point R are $(4,-2a)$

Circumcenter $△PQR=\frac{4+2t_1^2+2t_2^2}{2},\frac{-2a+4t_1+4t_2}{2}=(a^2-2,-3a)$

Hence, solved.