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: Understanding the Concept:
We need to find the specific coordinates of points \( P, Q, \) and \( R \) based on the slope conditions provided for each conic section. Once the coordinates are found, we can identify the geometric relationship between them. Often in such problems, the points form a right-angled triangle, which simplifies finding the circumradius significantly.
Step 2: Key Formula or Approach:
For any curve \( f(x, y) = 0 \), the slope of the tangent is \( m = \frac{dy}{dx} \).
Circumradius of a right triangle is half of its hypotenuse.
Distance formula: \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
Step 3: Detailed Explanation:
1. Point \( P \):
Parabola: \( y = x^2 \implies \frac{dy}{dx} = 2x \).
Given slope \( m = 4 \implies 2x = 4 \implies x = 2 \).
Then \( y = 2^2 = 4 \).
So, \( P = (2, 4) \).
2. Point \( Q \):
Circle: \( x^2 + y^2 = 2 \). Differentiating implicitly: \( 2x + 2y y' = 0 \implies y' = -x/y \).
Given slope \( m = -1 \implies -x/y = -1 \implies x = y \).
Substitute into circle: \( x^2 + x^2 = 2 \implies 2x^2 = 2 \implies x = 1 \) (since \( Q \) is in 1st quadrant).
So, \( Q = (1, 1) \).
3. Point \( R \):
Ellipse: \( x^2 + 4y^2 = 8 \). Differentiating implicitly: \( 2x + 8y y' = 0 \implies y' = -x/4y \).
Given slope \( m = -1/2 \implies -x/4y = -1/2 \implies 2x = 4y \implies x = 2y \).
Substitute into ellipse: \( (2y)^2 + 4y^2 = 8 \implies 4y^2 + 4y^2 = 8 \implies 8y^2 = 8 \implies y = 1 \).
Then \( x = 2(1) = 2 \).
So, \( R = (2, 1) \).
4. Geometry of Triangle \( PQR \):
Coordinates: \( P(2, 4), Q(1, 1), R(2, 1) \).
Observe that \( P \) and \( R \) share the same x-coordinate (\( x=2 \)), meaning the side \( PR \) is vertical.
Observe that \( Q \) and \( R \) share the same y-coordinate (\( y=1 \)), meaning the side \( QR \) is horizontal.
Thus, \( \triangle PQR \) is a right-angled triangle at vertex \( R(2, 1) \).
5. Calculation of Radius:
The hypotenuse of this right triangle is \( PQ \).
Length of \( PQ = \sqrt{(2-1)^2 + (4-1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10} \).
The radius of the circumcircle \( r = \frac{1}{2} \text{hypotenuse} = \frac{\sqrt{10}}{2} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}} \).
Step 4: Final Answer:
The radius is \( \sqrt{5/2} \).