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To determine the equations of the two lines that pass through the point [tex]\((-4, -5)\)[/tex] and are tangent to the graph of [tex]\(y = x^2 + 1\)[/tex], we need to follow these steps:
1. Identify the slope of the tangent line: For any point [tex]\( (a, a^2 + 1) \)[/tex] on the curve [tex]\( y = x^2 + 1 \)[/tex], the slope of the tangent line is given by the derivative of the function. The derivative of [tex]\( y = x^2 + 1 \)[/tex] is [tex]\( y' = 2x \)[/tex]. So, the slope of the tangent line at the point [tex]\( (a, a^2 + 1) \)[/tex] is [tex]\( 2a \)[/tex].
2. Equation of the tangent line: The equation of the tangent line at any point [tex]\( (a, a^2 + 1) \)[/tex] can be written as:
[tex]\[ y - (a^2 + 1) = 2a(x - a) \][/tex]
Simplifying, this becomes:
[tex]\[ y = 2a(x - a) + a^2 + 1 \][/tex]
3. Using the given point [tex]\((-4, -5)\)[/tex] to find [tex]\(a\)[/tex]: Since the tangent line must pass through [tex]\((-4, -5)\)[/tex], we substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -5 \)[/tex] into the tangent line equation:
[tex]\[ -5 = 2a(-4 - a) + a^2 + 1 \][/tex]
Simplify the equation:
[tex]\[ -5 = 2a(-4 - a) + a^2 + 1 \][/tex]
[tex]\[ -5 = -8a - 2a^2 + a^2 + 1 \][/tex]
[tex]\[ -5 = -8a - a^2 + 1 \][/tex]
[tex]\[ -6 = -8a - a^2 \][/tex]
[tex]\[ a^2 + 8a - 6 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], we get:
[tex]\[ a = \frac{-8 \pm \sqrt{64 + 24}}{2} \][/tex]
[tex]\[ a = \frac{-8 \pm \sqrt{88}}{2} \][/tex]
[tex]\[ a = \frac{-8 \pm 2\sqrt{22}}{2} \][/tex]
[tex]\[ a = -4 \pm \sqrt{22} \][/tex]
4. Finding the slopes at the points of tangency: Substitute each [tex]\(a\)[/tex] back into the formula for the slope [tex]\(2a\)[/tex].
- For [tex]\(a = -4 + \sqrt{22}\)[/tex]:
[tex]\[ \text{slope} = 2(-4 + \sqrt{22}) = -8 + 2\sqrt{22} \][/tex]
- For [tex]\(a = -4 - \sqrt{22}\)[/tex]:
[tex]\[ \text{slope} = 2(-4 - \sqrt{22}) = -8 - 2\sqrt{22} \][/tex]
5. Constructing the tangent line equations: We have the slopes, and now we construct the tangent line equations using the point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex].
The point-slope form for the lines passing through [tex]\((-4, -5)\)[/tex] is:
[tex]\[ y + 5 = m(x + 4) \][/tex]
- For [tex]\( m = -8 + 2\sqrt{22} \)[/tex]:
[tex]\[ y + 5 = (-8 + 2\sqrt{22})(x + 4) \][/tex]
[tex]\[ y = (-8 + 2\sqrt{22})(x + 4) - 5 \][/tex]
- For [tex]\( m = -8 - 2\sqrt{22} \)[/tex]:
[tex]\[ y + 5 = (-8 - 2\sqrt{22})(x + 4) \][/tex]
[tex]\[ y = (-8 - 2\sqrt{22})(x + 4) - 5 \][/tex]
Thus, the equations of the two lines passing through the point [tex]\((-4, -5)\)[/tex] and tangent to [tex]\( y = x^2 + 1 \)[/tex] are:
[tex]\[ y = (-8 + 2\sqrt{22})(x + 4) - 5 \][/tex]
[tex]\[ y = (-8 - 2\sqrt{22})(x + 4) - 5 \][/tex]
1. Identify the slope of the tangent line: For any point [tex]\( (a, a^2 + 1) \)[/tex] on the curve [tex]\( y = x^2 + 1 \)[/tex], the slope of the tangent line is given by the derivative of the function. The derivative of [tex]\( y = x^2 + 1 \)[/tex] is [tex]\( y' = 2x \)[/tex]. So, the slope of the tangent line at the point [tex]\( (a, a^2 + 1) \)[/tex] is [tex]\( 2a \)[/tex].
2. Equation of the tangent line: The equation of the tangent line at any point [tex]\( (a, a^2 + 1) \)[/tex] can be written as:
[tex]\[ y - (a^2 + 1) = 2a(x - a) \][/tex]
Simplifying, this becomes:
[tex]\[ y = 2a(x - a) + a^2 + 1 \][/tex]
3. Using the given point [tex]\((-4, -5)\)[/tex] to find [tex]\(a\)[/tex]: Since the tangent line must pass through [tex]\((-4, -5)\)[/tex], we substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -5 \)[/tex] into the tangent line equation:
[tex]\[ -5 = 2a(-4 - a) + a^2 + 1 \][/tex]
Simplify the equation:
[tex]\[ -5 = 2a(-4 - a) + a^2 + 1 \][/tex]
[tex]\[ -5 = -8a - 2a^2 + a^2 + 1 \][/tex]
[tex]\[ -5 = -8a - a^2 + 1 \][/tex]
[tex]\[ -6 = -8a - a^2 \][/tex]
[tex]\[ a^2 + 8a - 6 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], we get:
[tex]\[ a = \frac{-8 \pm \sqrt{64 + 24}}{2} \][/tex]
[tex]\[ a = \frac{-8 \pm \sqrt{88}}{2} \][/tex]
[tex]\[ a = \frac{-8 \pm 2\sqrt{22}}{2} \][/tex]
[tex]\[ a = -4 \pm \sqrt{22} \][/tex]
4. Finding the slopes at the points of tangency: Substitute each [tex]\(a\)[/tex] back into the formula for the slope [tex]\(2a\)[/tex].
- For [tex]\(a = -4 + \sqrt{22}\)[/tex]:
[tex]\[ \text{slope} = 2(-4 + \sqrt{22}) = -8 + 2\sqrt{22} \][/tex]
- For [tex]\(a = -4 - \sqrt{22}\)[/tex]:
[tex]\[ \text{slope} = 2(-4 - \sqrt{22}) = -8 - 2\sqrt{22} \][/tex]
5. Constructing the tangent line equations: We have the slopes, and now we construct the tangent line equations using the point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex].
The point-slope form for the lines passing through [tex]\((-4, -5)\)[/tex] is:
[tex]\[ y + 5 = m(x + 4) \][/tex]
- For [tex]\( m = -8 + 2\sqrt{22} \)[/tex]:
[tex]\[ y + 5 = (-8 + 2\sqrt{22})(x + 4) \][/tex]
[tex]\[ y = (-8 + 2\sqrt{22})(x + 4) - 5 \][/tex]
- For [tex]\( m = -8 - 2\sqrt{22} \)[/tex]:
[tex]\[ y + 5 = (-8 - 2\sqrt{22})(x + 4) \][/tex]
[tex]\[ y = (-8 - 2\sqrt{22})(x + 4) - 5 \][/tex]
Thus, the equations of the two lines passing through the point [tex]\((-4, -5)\)[/tex] and tangent to [tex]\( y = x^2 + 1 \)[/tex] are:
[tex]\[ y = (-8 + 2\sqrt{22})(x + 4) - 5 \][/tex]
[tex]\[ y = (-8 - 2\sqrt{22})(x + 4) - 5 \][/tex]
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