Chapter 4.2, Problem 65E

a.

To determine

Toprove:That the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(x_1,y_1\right)$ is $\frac{x_{1}x}{a^2}+\frac{y_{1}y}{b^2}=1$ .

Explanation of Solution

Given information:The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the point is $\left(x_1,y_1\right)$ .

Formula used:

Two-point form of line is

  $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\times \left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)\text{are}\text{the}\text{two}\text{points}\text{on}\text{the}\text{line}$.

Also product, quotient and chain rule of derivatives are used.

Proof:

The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

  $\begin{array}{l}\text{Let}\text{A}\left(x_1,y_1\right)\text{and}\text{B}\left(x_2,y_2\right)\text{be}\text{two}\text{points}\text{on}\text{the}\text{ellipse},\text{then}\\ \frac{x_1{}^2}{a^2}+\frac{y_1{}^2}{b^2}=1\mathrm{.....}\left(1\right)\\ \frac{x_2{}^2}{a^2}+\frac{y_2{}^2}{b^2}=1\mathrm{.....}\left(2\right)\end{array}$

Subtract equation (1) from equation (2)

  $\begin{array}{l}\frac{x_2{}^2-x_1{}^2}{a^2}+\frac{y_2{}^2-y_1{}^2}{b^2}=0\\ \frac{(x_2-x_1)(x_2+x_1)}{a^2}+\frac{(y_2-y_1)(y_2+y_1)}{b^2}=0\\ \frac{(y_2-y_1)(y_2+y_1)}{b^2}=-\frac{(x_2-x_1)(x_2+x_1)}{a^2}\\ \frac{(y_2-y_1)}{(x_2-x_1)}=-\frac{b^2}{a^2}\times \frac{(x_2+x_1)}{(y_2+y_1)}\mathrm{.......}(3)\end{array}$

The equation for the line AB in two-point form is:

  $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\times \left(x-x_1\right)\mathrm{.........}\left(4\right),\text{where}\text{A}\left(x_1,y_1\right)\text{and}\text{B}\left(x_2,y_2\right)\text{are}\text{the}\text{two}\text{points}\text{on}\text{the}\text{line}$

By using the result of equation (3) in equation (4)

  $\begin{array}{l}\left(y-y_1\right)=-\frac{b^2}{a^2}\times \frac{(x_2+x_1)}{(y_2+y_1)}\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(y_2+y_1)}{b^2}=-\frac{1}{a^2}\times (x_2+x_1)\times \left(x-x_1\right)\mathrm{.......}\left(5\right)\text{,}\text{this}\text{is}\text{the}\text{equation}\text{of}\text{line}\text{AB}\text{.}\end{array}$

The line AB becomes tangent to ellipse, if $\text{A}\left(x_1,y_1\right)\text{=}\text{B}\left(x_2,y_2\right)$

Then $x_1=x_2\text{and}{\text{y}}_1=y_2$ ................(6)

Put equation (6) in equation (5), we get

  $\begin{array}{l}\frac{\left(y-y_1\right)(y_1+y_1)}{b^2}=-\frac{1}{a^2}\times (x_1+x_1)\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(2y_1)}{b^2}=-\frac{1}{a^2}\times (2x_1)\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(y_1)}{b^2}=-\frac{(x_1)\times \left(x-x_1\right)}{a^2}\\ \frac{\left(y-y_1\right)(y_1)}{b^2}+\frac{(x_1)\times \left(x-x_1\right)}{a^2}=0\\ \frac{y\times y_1}{b^2}+\frac{x\times x_1}{a^2}-\left(\frac{y_1{}^2}{b^2}+\frac{x_1{}^2}{a^2}\right)=0\\ \text{by}\text{using}\text{equation}\left(1\right)\text{,}\text{we}\text{get}\\ \frac{y{y}_1}{b^2}+\frac{x{x}_1}{a^2}-1=0\\ \frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1\end{array}$

Here $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ is the equation of tangent AB.

Hence proved that the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(x_1,y_1\right)$ is $\frac{x_{1}x}{a^2}+\frac{y_{1}y}{b^2}=1$ .

b.

To determine

To calculate:The equation of tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $\left(x_1,y_1\right)$ .

Answer to Problem 65E

The tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $\left(x_1,y_1\right)$ is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ .

Explanation of Solution

Given information:The hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the point is $\left(x_1,y_1\right)$ .

Formula used:

Two-point form of line is

  $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\times \left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)\text{are}\text{the}\text{two}\text{points}\text{on}\text{the}\text{line}$.

Also product, quotient and chain rule of derivatives are used.

Calculation:

The hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

  $\begin{array}{l}\text{Let}\text{A}\left(x_1,y_1\right)\text{and}\text{B}\left(x_2,y_2\right)\text{be}\text{two}\text{points}\text{on}\text{the}\text{hyperbola},\text{then}\\ \frac{x_1{}^2}{a^2}-\frac{y_1{}^2}{b^2}=1\mathrm{.....}\left(1\right)\\ \frac{x_2{}^2}{a^2}-\frac{y_2{}^2}{b^2}=1\mathrm{.....}\left(2\right)\end{array}$

Subtract equation (1) from equation (2)

  $\begin{array}{l}\frac{x_2{}^2-x_1{}^2}{a^2}-\frac{y_2{}^2-y_1{}^2}{b^2}=0\\ \frac{(x_2-x_1)(x_2+x_1)}{a^2}-\frac{(y_2-y_1)(y_2+y_1)}{b^2}=0\\ \frac{(y_2-y_1)(y_2+y_1)}{b^2}=\frac{(x_2-x_1)(x_2+x_1)}{a^2}\\ \frac{(y_2-y_1)}{(x_2-x_1)}=\frac{b^2}{a^2}\times \frac{(x_2+x_1)}{(y_2+y_1)}\mathrm{.......}(3)\end{array}$

The equation for the line AB in two-point form is:

  $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\times \left(x-x_1\right)\mathrm{.........}\left(4\right),\text{where}\text{A}\left(x_1,y_1\right)\text{and}\text{B}\left(x_2,y_2\right)\text{are}\text{the}\text{two}\text{points}\text{on}\text{the}\text{line}$

By using the result of equation (3) in equation (4)

  $\begin{array}{l}\left(y-y_1\right)=\frac{b^2}{a^2}\times \frac{(x_2+x_1)}{(y_2+y_1)}\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(y_2+y_1)}{b^2}=\frac{1}{a^2}\times (x_2+x_1)\times \left(x-x_1\right)\mathrm{.......}\left(5\right)\text{,}\text{this}\text{is}\text{the}\text{equation}\text{of}\text{line}\text{AB}\text{.}\end{array}$

The line AB becomes tangent to hyperbola, if $\text{A}\left(x_1,y_1\right)\text{=}\text{B}\left(x_2,y_2\right)$

Then $x_1=x_2\text{and}{\text{y}}_1=y_2$ ................(6)

Put equation (6) in equation (5), we get

  $\begin{array}{l}\frac{\left(y-y_1\right)(y_1+y_1)}{b^2}=\frac{1}{a^2}\times (x_1+x_1)\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(2y_1)}{b^2}=\frac{1}{a^2}\times (2x_1)\times \left(x-x_1\right)\\ \frac{\left(y-y_1\right)(y_1)}{b^2}=\frac{(x_1)\times \left(x-x_1\right)}{a^2}\\ \frac{(x_1)\times \left(x-x_1\right)}{a^2}-\frac{\left(y-y_1\right)(y_1)}{b^2}=0\\ \frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-\left(\frac{x_1{}^2}{a^2}-\frac{y_1{}^2}{b^2}\right)=0\\ \text{by}\text{using}\text{equation}\left(1\right)\text{,}\text{we}\text{get}\\ \frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-1=0\\ \frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1\end{array}$

Here $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ is the equation of tangent AB.

Hence the tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $\left(x_1,y_1\right)$ is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ .