Question

Finding Tangents (a) Show that the tangent to the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ at the point \(\left(x_{1}, y_{1}\right)\) has equation $$\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$$ (b) Find an equation for the tangent to the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ at the point \(\left(x_{1}, y_{1}\right)\)

Short answer

The equation of the tangent to the ellipse at the point \((x_{1}, y_{1})\) is indeed \(\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1\) as derivable from the equation of the ellipse. The equation of the tangent to the hyperbola at any point \((x_{1}, y_{1})\) can similarly be derived from the equation of the hyperbola by finding the derivative, hence obtaining the slope of the tangent line, and then using this to find the equation of the tangent line.

Step-by-step solution

1. Derive the Equation of the Tangent for the Ellipse

Take the equation of the ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and differentiate it with respect to x to get the gradient of the tangent. So, we have \(\frac{2x}{a^{2}}+\frac{2y}{b^{2}} y' = 0\). Solving for y', we get \( y' = -\frac{x \cdot b^{2}}{y \cdot a^{2}} \). The equation of the line at \((x_{1}, y_{1})\) will be \(y - y_{1} = y'(x - x_{1})\). Substituting \(y'\) and simplifying, we get \(\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1\) as required.

2. Derive the Equation of the Tangent for the Hyperbola

Proceed in the same way as for the ellipse. Take the equation of the hyperbola, \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), and differentiate it with respect to x to find the gradient of the tangent line. So, we get \(\frac{2x}{a^{2}}-\frac{2y}{b^{2}} y' = 0\). Solving for y', we have \( y' = \frac{x \cdot b^{2}}{y \cdot a^{2}}\). Substituting this into the equation of the line \(y - y_{1} = y'(x - x_{1})\), and simplifying, the equation for the tangent line to the hyperbola is obtained.