Find the equations of tangent and normal to the parabola x2 6x 4y 5=0 at (1,-3)?

Question

Answer

Equations of Tangent and Normal to the Parabola

Given equation of the parabola is:

x² + 6x + 4y + 5 = 0

Step 1: Find the Derivative of the Parabola

To find the equation of the tangent and normal at a point on the parabola, we need to first find the derivative of the parabola. The derivative of the given equation is:

dy/dx = (-2x - 6)/4 = -x/2 - 3/2

Step 2: Find the Slope of the Tangent

Now, we can find the slope of the tangent at the point (1,-3) by plugging in x = 1 into the derivative:

dy/dx = -1/2 - 3/2 = -2

Step 3: Find the Equation of the Tangent

The equation of the tangent at (1,-3) can be found using the point-slope form:

y - y₁ = m(x - x₁)

Substituting the values, we get:

y + 3 = -2(x - 1)

y = -2x - 5

Step 4: Find the Slope of the Normal

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal at (1,-3) is:

m = 1/2

Step 5: Find the Equation of the Normal

Using the point-slope form, the equation of the normal can be found:

y - y₁ = m(x - x₁)

Substituting the values, we get:

y + 3 = 1/2(x - 1)

y = 1/2x - 5/2

Conclusion

Therefore, the equation of the tangent at (1,-3) is y = -2x - 5 and the equation of the normal is y = 1/2x - 5/2.

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