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Find the equation of the parabola whose equation of the tangent at vertex is
3x - 4y = 5 and focus at the point
(1, 2).

User Madelyn
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Answer: The equation of the tangent to a parabolic curve y = ax^2 + bx + c at the point (h, k) can be expressed as:

y = a(x - h)^2 + k

So, we know the equation of the tangent at vertex of the parabolic curve is 3x - 4y = 5, so the vertex of the parabolic curve is (h, k) = (2, -3/2).

The equation of a parabolic curve with vertex (h, k) and focus (h, k + p) is given by:

y = a(x - h)^2 + k, where a = -1/4p.

So, substituting the values of the vertex (2, -3/2) and the focus (1, 2) into the above equation, we have:

-3/2 = a(2 - 2)^2 - 3/2, so a = -1/4p.

And, substituting the values of the focus (1, 2) into the equation y = a(x - h)^2 + k, we have:

2 = -1/4p(1 - 2)^2 - 3/2, so p = 1/2.

So, the equation of the parabolic curve with focus (1, 2) and vertex (2, -3/2) is:

y = -1/4 * 1/2 * (x - 2)^2 - 3/2.

This is the equation of the parabola that has the equation of the tangent at vertex 3x - 4y = 5 and focus at the point (1, 2).

Explanation:

User Msquared
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