Parabola

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Question

$If the normals to the curve y = x^{2} at the points P, Q and R nass through the point (0, 3 / 2) then the radius of the$
$circle circumscribing Δ P Q R is$

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Solution

$Equation of normal at any point (t, t^{2}) is$
$\begin{array}{l} y - t^{2} = - \frac{1}{2 t} (x - t) \\ It passes through (0, \frac{3}{2}) \\ ∴ t = 0 or t = 1, - 1 \end{array}$
$Hence P, Q, R are (0, 0) (1, 1) and (- 1, 1) .$
$\Rightarrow P Q R is a right triangle.$
$∴ radius of circumcircle is 1 .$

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Parabola

Remember concepts with our Masterclasses.

If the normals to the curve y=x2 at the points P,Q and R nass through the point (0,3/2) then the radius of the circle circumscribing ΔPQR is

Equation of normal at any point t,t2 is y−t2=−12t(x−t) It passes through 0,32∴t=0 or t=1,−1 Hence P,Q,R are (0,0)(1,1) and (−1,1) . ⇒PQR is a right triangle. ∴ radius of circumcircle is 1.

Ready to Test Your Skills?

free study material

$If the normals to the curve y = x^{2} at the points P, Q and R nass through the point (0, 3 / 2) then the radius of the$
$circle circumscribing Δ P Q R is$