Let's intersect them!(For JEE ADV 2016)

Geometry Level 4

Consider the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$ and the parabola ${ y }^{ 2 }=8x$ . They intersect at $P$ and $Q$ in the first and fourth quadrant respectively. The tangents to the circle at $P$ and $Q$ intersect the $x$ -axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $x$ axis at $S$ . Then the ratio of areas of $\Delta PQS$ and $\Delta PQR$ is $a:b$ , where $a$ and $b$ are coprime positive integers. Find $a+b$ .

The answer is 5.

1 solution

Niranjan Khanderia
Jul 8, 2016

Let PQ interest SR at M, and O(0,0) be the center of the circle.
P is on both curves, so y coordinates as on each must equal.
$\implies\ x^2+8x=9,\ \ \therefore\ (x+9)(x-1)=0.\ \text{ x can not be -9, so x=1.}\\ So\ P(1,2\sqrt2), \ SM=2\sqrt2.$ .
With common angle ORP, right triangles OPR and PMR are similar.
$\therefore\ PM^2=OM*MR,\ \implies\ 8=1*MR,................................. MR=8.\\$
Slope of PS equals derivative of parabola at P.
$\dfrac {dy}{dx}=\dfrac 8 {y_p}=\dfrac8{2\sqrt2}=\sqrt2\ but =TanPSR\ also\ =\dfrac{PM}{SM}\\ So\ SM=2.$ .
Triangles PQS and PQR have common base, so their areas are proportional to their heights.
Ratio of areas is SM : MR :: 2 : 8 :: 1 : 4. So 1+4= $\huge\ \ \color{#D61F06}{**5**}$ .

Let's intersect them!(For JEE ADV 2016)

The answer is 5.

1 solution

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