Finding the Positive Number A in a Parabola Problem

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Homework Help Overview

The problem involves finding a positive number A related to the tangents of the parabola defined by the equation y=x²+3, which pass through the point (0,–2). The tangents touch the parabola at points (A, A²+3) and (–A, A²+3).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to start the problem and seeks step-by-step guidance. Some participants discuss the slope of the tangent line and its derivation, while others question the notation used and the steps to find A.

Discussion Status

Participants are exploring the relationship between the slope of the tangent line and the points of tangency on the parabola. Some guidance has been provided regarding the formulation of the tangent line and its slope, but there is no explicit consensus on the next steps or the value of A.

Contextual Notes

There is a mention of different notations for mathematical expressions, and some participants express uncertainty about the derivation of the slope and the solution for A.

i3uddha
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alright so here is the problem.
The parabola y=x**2+3 has two tangents which pass through the point (0,–2). One is tangent to the to the parabola at (A,A2+3) and the other at (–A,A2+3). Find (the positive number) A.

i've attempted this problem several times but haven't been able to find the solution. :confused:
could someone tell me step by step how i would start this problem?
 
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Better to use x^2 rather than x**2 (I haven't seen that notation since BASIC). Any line through (0, -2) can be written y= mx- 2. A line tangent to y= x^2+ 3 at (A, A^2+ 3) must have slope m= 2A: y= 2Ax- 2
Determine A so that y= A^2+ 3= 2A(A)- 2= 3A^2- 2.
 
how did you get the slope to be 2A?
and what would the answer for A be?
 
The derivative of y= x^2 is y'= 2x. At x= A, y'= 2A and that is the slope ofthe tangent line.

Surely you can solve A^2+ 3= 3A^2- 2 yourself.
 

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