Question

For the parabola $y^2=8x$, tangent and normal are drawn at $P(2,4)$ which meet the axis of the parabola in $A$ and $B$, then the length of the diameter of the circle through $A,P,B$ is

• A. $2$
• B. $4$
• C. $8$
• D. $6$

Answer 1

Answer: (C)

$8$

Considering the equation of the parabola

$y^2=8x$

The axis of symmetry is the positive $x-$axis.

By, differentiating with respect to $x$, we get

$2y.y^{\prime }=8$
$y^{\prime }=\frac{4}{y}$

Hence, $y_{(2,4)}^{\prime }=1$

Thus slope of the tangent at the point of contact $P(2,4)$ is $1$ while that of normal is $-1$.

Hence, equation of the tangent will be

$\frac{y-4}{x-2}=1$
$-x+y=2$ ...(i)

Hence, the tangent meets the x axis at $(-2,0)$

Therefore, $A=(-2,0)$.

Equation of normal will be

$\frac{y-4}{x-2}=-1$
$x+y=6$ ...(ii)

Hence the normal meets the x axis at $(6,0)$.

Thus, $B=(6,0)$

Therefore the three points through which the circle passes are

$(-2,0),(6,0),(2,4)$

Now let the equation of the circle be

$(x-h{)}^2+(y-k{)}^2=r^2$

Therefore

$(-2-h{)}^2+k^2=r^2$

$\to (2+h{)}^2+k^2=r^2$
$(6-h{)}^2+k^2=r^2$

Subtracting {ii} from {i}, we get

$2h-4=0$
$h=2$

Therefore the equation of the circle reduces to

$(x-2{)}^2+(y-k{)}^2=r^2$
Now

$(2-2{)}^2+(4-k{)}^2=r^2$

$\to (4-k{)}^2=r^2$
$(6-2{)}^2+k^2=r^2$

$\to 16+k^2=r^2$

Subtracting i from ii, we get

$16+k^2-(k-4{)}^2=0$
$16+(2k-4)\left(4\right)=0$
$4+2k-4=0$
$k=0$

Thus the centre of the circle lies at $C=(2,0)$.

Hence the radius of the circle is

$CA=CP=CB$
Now

$CA=\sqrt{(2-(-2){)}^2+0^2}$

$=2-(-2)$
$=4$

$=r$

Hence, the diameter of the circle is $8$ units.