Conic Sections Visualizer

First, rewrite the ellipse equation in standard form: \[ \frac{x^2}{7} + \frac{y^2}{2} = 1 \]

The condition for a line \(y = mx + c\) to be tangent to this ellipse is: \[ c^2 = 7m^2 + 2 \]

Since the tangent passes through (5,2), substitute: \[ 2 = 5m + c \Rightarrow c = 2 - 5m \]

Substitute \(c\) into the tangency condition: \[ (2 - 5m)^2 = 7m^2 + 2 \] \[ 25m^2 - 20m + 4 = 7m^2 + 2 \] \[ 18m^2 - 20m + 2 = 0 \] \[ 9m^2 - 10m + 1 = 0 \]

Solve the quadratic equation: \[ m = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm 8}{18} \] So \(m_1 = 1\) and \(m_2 = \frac{1}{9}\)

First, find the slope of the given line \(10x - 3y + 9 = 0\): \[ 3y = 10x + 9 \Rightarrow y = \frac{10}{3}x + 3 \] So slope \(m = \frac{10}{3}\)

The condition for tangency to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is: \[ c^2 = a^2m^2 - b^2 \] Here \(a^2 = 16\), \(b^2 = 64\), and \(m = \frac{10}{3}\)

Calculate \(c\): \[ c^2 = 16 \left(\frac{10}{3}\right)^2 - 64 = \frac{1600}{9} - 64 = \frac{1600 - 576}{9} = \frac{1024}{9} \] \[ c = \pm \frac{32}{3} \]

Rewrite the line equation: \(y = x + 4\)

Substitute into the ellipse equation: \[ x^2 + 3(x + 4)^2 = 12 \] \[ x^2 + 3(x^2 + 8x + 16) = 12 \] \[ 4x^2 + 24x + 48 - 12 = 0 \] \[ 4x^2 + 24x + 36 = 0 \] \[ x^2 + 6x + 9 = 0 \]

The discriminant is: \[ D = 36 - 36 = 0 \] Since \(D = 0\), the line touches the ellipse at exactly one point.

Find the point of contact by solving: \[ x^2 + 6x + 9 = 0 \Rightarrow (x + 3)^2 = 0 \Rightarrow x = -3 \] Then \(y = -3 + 4 = 1\)

Find slope of given line \(2x + 2y + 3 = 0\): \[ y = -x - \frac{3}{2} \Rightarrow m = -1 \]

Slope of perpendicular line is negative reciprocal: \[ m_{\text{tangent}} = 1 \]

For parabola \(y^2 = 4ax\), equation of tangent with slope \(m\) is: \[ y = mx + \frac{a}{m} \] Here \(4a = 16 \Rightarrow a = 4\)

Substitute \(m = 1\) and \(a = 4\): \[ y = x + \frac{4}{1} \Rightarrow y = x + 4 \]

The parametric form of parabola \(y^2 = 4ax\) is: \[ x = at^2, y = 2at \] Here \(4a = 8 \Rightarrow a = 2\)

At \(t = 2\), the point is: \[ x = 2(2)^2 = 8 \] \[ y = 2(2)(2) = 8 \] So point is \((8, 8)\)

The tangent equation in parametric form is: \[ ty = x + at^2 \]

Substitute \(t = 2\) and \(a = 2\): \[ 2y = x + 2(2)^2 \] \[ 2y = x + 8 \]

First rewrite the hyperbola in standard form: \[ \frac{x^2}{9} - \frac{y^2}{12} = 1 \] Here \( a = 3 \), \( b = 2\sqrt{3} \)

The parametric equations are: \[ x = a\sec\theta = 3\sec\left(\frac{\pi}{3}\right) = 6 \] \[ y = b\tan\theta = 2\sqrt{3}\tan\left(\frac{\pi}{3}\right) = 6 \]

The tangent equation is: \[ \frac{x\sec\theta}{a} - \frac{y\tan\theta}{b} = 1 \] \[ \frac{6x}{9} - \frac{6y}{12} = 1 \]

Simplify the tangent: \[ \frac{2x}{3} - \frac{y}{2} = 1 \] \[ 4x - 3y - 6 = 0 \]

The normal equation is: \[ \frac{ax}{\sec\theta} + \frac{by}{\tan\theta} = a^2 + b^2 \] \[ \frac{3x}{2} + \frac{2\sqrt{3}y}{\sqrt{3}} = 9 + 12 \]

The tangent at \( t_1 \) is: \[ yt_1 = x + at_1^2 \]

The tangent at \( t_2 \) is: \[ yt_2 = x + at_2^2 \]

Subtract the two equations: \[ y(t_1 - t_2) = a(t_1^2 - t_2^2) \] \[ y = a(t_1 + t_2) \]

Substitute back to find x: \[ x = yt_1 - at_1^2 = at_1(t_1 + t_2) - at_1^2 = at_1t_2 \]

The normal at \( t_1 \) is: \[ y = -t_1x + 2at_1 + at_1^3 \]

This normal meets the parabola again at \( t_2 \), so: \[ at_2^2 = -t_1(2at_2) + 2at_1 + at_1^3 \]

Simplify the equation: \[ t_2^2 + 2t_1t_2 - 2t_1 - t_1^3 = 0 \]

We know \( t_2 \neq t_1 \), so divide by \( (t_2 - t_1) \): \[ t_2 = -t_1 - \frac{2}{t_1} \]