20:[["$","title","0",{"children":"BMSAT Kozhikode 2025 - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined A point on the curve y = x^2 has coordinates (x, 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)

(x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

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 y = 4 Therefore, 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"}],["$","meta","3",{"name":"keywords","content":"BMSAT Kozhikode,QA,Geometry,Conic Sections"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"name":"mobile-web-app-capable","content":"yes"}],["$","meta","8",{"name":"apple-mobile-web-app-title","content":"AfterBoards"}],["$","meta","9",{"name":"apple-mobile-web-app-status-bar-style","content":"default"}],["$","meta","10",{"property":"og:title","content":"BMSAT Kozhikode 2025 - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions"}],["$","meta","11",{"property":"og:description","content":"undefined A point on the curve y = x^2 has coordinates (x, 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)

(x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

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 y = 4 Therefore, the point (2, 4) on the curve y = x^2 is equidistant from (-1, 0) and (5, 0)."}],["$","meta","12",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/3"}],["$","meta","13",{"property":"og:site_name","content":"AfterBoards | BMSAT Kozhikode 2025 Preparation"}],["$","meta","14",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","15",{"property":"og:type","content":"website"}],["$","meta","16",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","17",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","18",{"name":"twitter:title","content":"BMSAT Kozhikode 2025 - Consider the statement: \"A point on the curve y = x² is equidistant from the points (-1,0) and (5,0).\" | PYQs + Solutions"}],["$","meta","19",{"name":"twitter:description","content":"undefined A point on the curve y = x^2 has coordinates (x, 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)

(x + 1)^2 + (x^2)^2 = (x - 5)^2 + (x^2)^2 The (x^2)^2 terms cancel: (x + 1)^2 = (x - 5)^2

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 y = 4 Therefore, the point (2, 4) on the curve y = x^2 is equidistant from (-1, 0) and (5, 0)."}],"$L2e","$L2f","$L30","$L31","$L32","$L33","$L34","$L35","$L36","$L37"]

Consider the statement: "A point on the curve $y = x²$ is equidistant from the points $(-1,0)$ and $(5,0).$"

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