Question

Let \(P\) be the point on the parabola \(y=x^2\) such that the slope of the tangent to the parabola at the point \(P\) is \(4\). Let \(Q\) be the point in the first quadrant lying on the circle \[ x^2+y^2=2 \] such that the slope of the tangent to the circle at the point \(Q\) is \(-1\). Let \(R\) be the point in the first quadrant lying on the ellipse \[ x^2+4y^2=8 \] such that the slope of the tangent to the ellipse at the point \(R\) is \(-\frac12\). Then the radius of the circle passing through the points \(P\), \(Q\) and \(R\) is:

• A. \(\sqrt{10}\)
• B. \(\sqrt5\)
• C. \(\sqrt{\frac52}\)
• D. \(2\sqrt5\)

Hint:

For a right triangle: \[ \mathrm{Circumradius} = \frac{\mathrm{Hypotenuse}}{2} \]

Answer 1

The Correct Option is C

Step 1: Find point \(P\).
For parabola: \[ y=x^2 \] Slope of tangent: \[ \frac{dy}{dx}=2x \] Given: \[ 2x=4 \] \[ x=2 \] Thus: \[ y=2^2=4 \] Hence: \[ P=(2,4) \]

Step 2: Find point \(Q\).
Given circle: \[ x^2+y^2=2 \] Differentiating: \[ 2x+2y\frac{dy}{dx}=0 \] \[ \frac{dy}{dx}=-\frac{x}{y} \] Given slope: \[ -\frac{x}{y}=-1 \] \[ x=y \] Substituting in: \[ x^2+y^2=2 \] \[ 2x^2=2 \] \[ x=1 \] Since first quadrant: \[ y=1 \] Thus: \[ Q=(1,1) \]

Step 3: Find point \(R\).
Given ellipse: \[ x^2+4y^2=8 \] Differentiating: \[ 2x+8y\frac{dy}{dx}=0 \] \[ \frac{dy}{dx}=-\frac{x}{4y} \] Given: \[ -\frac{x}{4y}=-\frac12 \] \[ x=2y \] Substituting: \[ (2y)^2+4y^2=8 \] \[ 4y^2+4y^2=8 \] \[ 8y^2=8 \] \[ y=1 \] Thus: \[ x=2 \] Hence: \[ R=(2,1) \]

Step 4: Find the circumradius of triangle \(PQR\).
Coordinates: \[ P=(2,4),\quad Q=(1,1),\quad R=(2,1) \] Side lengths: \[ QR=1 \] \[ PR=3 \] \[ PQ=\sqrt{(2-1)^2+(4-1)^2} \] \[ =\sqrt{10} \] Triangle \(PQR\) is right angled at \(R\). Circumradius of right triangle: \[ =\frac{\text{Hypotenuse}}{2} \] Thus: \[ \text{Radius}=\frac{\sqrt{10}}{2} \] \[ =\sqrt{\frac52} \]

Step 5: Identify the correct option.
Therefore: \[ \boxed{\mathrm{(C)\ }\sqrt{\frac52}} \]