Question

Let $PQ$ be a chord of the parabola $y^2=4x$. A circle drawn with $PQ$ as diameter passes through the vertex ${}^{\prime }{V}^{\prime }$ of the parabola. If the area of triangle $PVQ$ is $20$ then coordinates of $P$ are

• A. $(-16,-18)$
• B. $(-16,8)$
• C. $(16,-8)$
• D. $(8,16)$

Answer 1

The correct option is C $(16,-8)$

R.E.F image

Slope of PV:-

$\frac{y_2-y_1}{x_2-x_1}=\frac{2t-0}{t^2-0}=\frac{2t}{t^2}=\frac{2}{t}$

The angle sustained on diameter is always ${90}^0$

So, $PV$ and $VQ$ are perpendicular to each other.

So, slope of $VQ=\frac{-t}{2}$ $[\therefore slopePV\times VQ=-1]$

equation of $VQ,y=\frac{-tx}{2}$.........(1)

$VQ$ is interest by parabola

So, find coordinates of $Q$,

by solving parabola at the line $VQ$

Parabola $\to y^2=4x$

$=\frac{t^2{x}^2}{4}=4x$.........From(1)

$x=\frac{16}{t^2}$

and $y^2=4\times \frac{16}{t^2}\Rightarrow y=\frac{8}{t}$

So, $Q(\frac{16}{t^2},\frac{8}{t})$

Now, $Ar\left(PVQ\right)=20$

$\Rightarrow \frac{1}{2}\times PV\times VQ=20\Rightarrow PV\times VQ=40$

$\Rightarrow \sqrt{(t^2-0{)}^2+(2t-0{)}^2}\times \sqrt{(\frac{t^6}{t^2}-0)+{(\frac{-8}{t}-0)}^2}=40$ [distance of $PQ$ and $QV$]

squaring both side we get-

$(t^4+4t^2)(\frac{256}{t^4}+\frac{64}{t^2})=1600$

$\Rightarrow 256+64t^2+\frac{256\times 4}{t^2}+256=1600$

$\Rightarrow 64t^2+\frac{256\times 4}{t^2}+5+2=1600$

$\Rightarrow 64t^2+\frac{256\times 4}{t^2}=1088$

$64t^4+256\times 4-1088t^2=0$

$\Rightarrow 64t^4-1088t^2+256\times 4=0$

$\Rightarrow t^4-17t^2+4\times 4=0$

$\Rightarrow t^4-17t^2+16=0$

$\Rightarrow t^4-16t^2-t^2+16=0$

$\Rightarrow t^2(t^2-16)-1(t^2-16)=0$

$\Rightarrow (t^2-1)(t^2-16)=0$

$t=\pm 1,t=\pm \sqrt{16},t=\pm 4$

$P(t^2,2t)$

$\Rightarrow P(1,2)$ or $p(1,-2)$ or $P(16,8)$, $P(16,-8)$

So, option $C$ is correct $(16,-8)$