d:[["$","title","0",{"children":"IIM Kozhikode - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at this prob<=m, we need to find a point on the curve y = x^2 that's the same distance from both (-1,0) and (5,0). Let's call the point on the curve (x, x^2) since any point on y = x^2 has coordinates where the y-value equals x^2.

Using the distance formula: Distance from (x, x^2) to (-1, 0): sqrt((x + 1)^2 + (x^2)^2) Distance from (x, x^2) to (5, 0): sqrt((x - 5)^2 + (x^2)^2)

Since the point is equidistant from both points: sqrt((x + 1)^2 + (x^2)^2) = sqrt((x - 5)^2 + (x^2)^2)

Squaring both sides to eliminate the square roots: (x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

Expanding both sides: x^2 + 2x + 1 = x^2 - 10x + 25 The x^2 terms cancel: 2x + 1 = -10x + 25 12x = 24 x = 2

Since the point is on y = x^2: y = (2)^2 = 4 The point (2, 4) on the curve y = x^2 is equidistant from (-1, 0) and (5, 0)."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Geometry,Conic Sections"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"IIM Kozhikode - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at this prob<=m, we need to find a point on the curve y = x^2 that's the same distance from both (-1,0) and (5,0). Let's call the point on the curve (x, x^2) since any point on y = x^2 has coordinates where the y-value equals x^2.

Using the distance formula: Distance from (x, x^2) to (-1, 0): sqrt((x + 1)^2 + (x^2)^2) Distance from (x, x^2) to (5, 0): sqrt((x - 5)^2 + (x^2)^2)

Since the point is equidistant from both points: sqrt((x + 1)^2 + (x^2)^2) = sqrt((x - 5)^2 + (x^2)^2)

Squaring both sides to eliminate the square roots: (x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

Expanding both sides: x^2 + 2x + 1 = x^2 - 10x + 25 The x^2 terms cancel: 2x + 1 = -10x + 25 12x = 24 x = 2

Since the point is on y = x^2: y = (2)^2 = 4 The point (2, 4) on the curve y = x^2 is equidistant from (-1, 0) and (5, 0)."}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/3"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | IIM Kozhikode Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"IIM Kozhikode - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at this prob<=m, we need to find a point on the curve y = x^2 that's the same distance from both (-1,0) and (5,0). Let's call the point on the curve (x, x^2) since any point on y = x^2 has coordinates where the y-value equals x^2.

Using the distance formula: Distance from (x, x^2) to (-1, 0): sqrt((x + 1)^2 + (x^2)^2) Distance from (x, x^2) to (5, 0): sqrt((x - 5)^2 + (x^2)^2)

Since the point is equidistant from both points: sqrt((x + 1)^2 + (x^2)^2) = sqrt((x - 5)^2 + (x^2)^2)

Squaring both sides to eliminate the square roots: (x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

Expanding both sides: x^2 + 2x + 1 = x^2 - 10x + 25 The x^2 terms cancel: 2x + 1 = -10x + 25 12x = 24 x = 2

Since the point is on y = x^2: y = (2)^2 = 4 The point (2, 4) on the curve y = x^2 is equidistant from (-1, 0) and (5, 0)."}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

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