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Theory of Equations: Vertex Quadratic Expression Quadratic Equation Whose Vertex

JEE Maths question with a full step-by-step solution.

Question
If the vertex of the quadratic expression y=2x2−4x+6y = 2x^2 - 4x + 6 is (m,n)(m, n), then the quadratic equation whose vertex is (2m,n2)\left(2m, \dfrac{n}{2}\right) is
Ay=x2−4x+8y = x^2 - 4x + 8
By=x2+4x+6y = x^2 + 4x + 6
Cy=−2x2+8x−6y = -2x^2 + 8x - 6correct
Dy=−2x2+8x−8y = -2x^2 + 8x - 8
Solution
Step 1: Find the vertex of y=2x2−4x+6y = 2x^2 - 4x + 6.
m=−b2a=44=1,n=−D4a=−16−488=4m = -\frac{b}{2a} = \frac{4}{4} = 1, \qquad n = -\frac{D}{4a} = -\frac{16-48}{8} = 4
 So (m,n)=(1,4)(m, n) = (1, 4). Step 2: The required vertex is (2m,n2)=(2,2)\left(2m, \dfrac{n}{2}\right) = (2, 2). Step 3: Verify option (C): y=−2x2+8x−6y = -2x^2 + 8x - 6. Vertex at x=−82⋅(−2)=2x = \dfrac{-8}{2 \cdot (-2)} = 2 and y=−2(4)+16−6=2y = -2(4)+16-6 = 2. Vertex is (2,2)(2, 2). ✓ Correct answer: (3)
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