Answer to Question #270950 in Analytic Geometry for Aliyah

Question

Find the equations of the tangent and normal to each of the following conics, the lengths of the subtangent and subnormal, then trace the curve showing these lines.

1. y2 = -9 (x – 3) at (2, 3).
2. 4x2 + 3y2 + 16x + 36y – 68 = 0 at (4, -2)
3. 9x2 – 4y2 – 108x – 56y + 128 = 0 at (0, 2).
4. x2 + y2 – 8x – 6y + 15 = 0 at (5, 0).

Answer 1

1.

$y^2=-9(x-3)$

$2yy'=-9$

$y'=\dfrac{-9}{2y}$

Point $(2, 3)$

$slope=m_1=y'=\dfrac{-9}{2(3)}=-\dfrac{3}{2}$

Tangent line is

$y-3=-\dfrac{3}{2}(x-2)$

The equation of the tangent line in slope-intercept form

$y=-\dfrac{3}{2}x+6$

Find the equation of the normal at the point $(2, 3)$

$slope=m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}$

$y-3=\dfrac{2}{3}(x-2)$

The equation of the normal line in slope-intercept form

$y=\dfrac{2}{3}x+\dfrac{5}{3}$

$TA$   is defined to be the subtangent at $P$. $AN$  is called the subnormal.

The lengths $PT$  and $PN$  are called the tangent and normal.

$A(2, 0), T(4, 0), N(-\dfrac{5}{2}, 0)$

$PT=\sqrt{(4-2)^2+(0-3)^2}=\sqrt{13}$

$PN=\sqrt{(-\dfrac{5}{2}-2)^2+(0-3)^2}=\dfrac{3}{2}\sqrt{13}$

$TA=2$

$AN=\dfrac{9}{2}$

2.

$4x^2 + 3y^2 + 16x + 36y-68 = 0$

$8x+6yy'+16+36y'=0$

$y'=-\dfrac{4x+8}{3y+18}$

Point $(4, -2)$

$slope=m_1=-\dfrac{4(4)+8}{3(-2)+18}=-2$

Tangent line is

$y-(-2)=-2(x-4)$

The equation of the tangent line in slope-intercept form

$y=-2x+6$

Find the equation of the normal at the point $(4, -2)$

$slope=m_2=-\dfrac{1}{m_1}=\dfrac{1}{2}$

$y-(-2)=\dfrac{1}{2}(x-4)$

The equation of the normal line in slope-intercept form

$y=\dfrac{1}{2}x-4$

$TA$   is defined to be the subtangent at $P$. $AN$  is called the subnormal.

The lengths $PT$  and $PN$  are called the tangent and normal.

$A(4, 0), T(3, 0), N(8, 0)$

$PT=\sqrt{(3-4)^2+(0-(-2))^2}=\sqrt{5}$

$PN=\sqrt{(8-4)^2+(0-(-2))^2}=2\sqrt{5}$

$TA=1$

$AN=4$

3.

$9x^2- 4y^2-108x- 56y+128 = 0$

$18x-8yy'-108-56y'=0$

$y'=\dfrac{9x-54}{4y+28}$

Point $(0, 2)$

$slope=m_1=\dfrac{9(0)-54}{4(2)+28}=-\dfrac{3}{2}$

Tangent line is

$y-2=-\dfrac{3}{2}(x-0)$

The equation of the tangent line in slope-intercept form

$y=-\dfrac{3}{2}x+2$

Find the equation of the normal at the point $(0, 2)$

$slope=m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}$

$y-2=\dfrac{2}{3}(x-0)$

The equation of the normal line in slope-intercept form

$y=\dfrac{2}{3}x+2$

$TA$   is defined to be the subtangent at $P$. $AN$  is called the subnormal.

The lengths $PT$  and $PN$  are called the tangent and normal.

$A(0, 0), T(\dfrac{4}{3}, 0), N(-3, 0)$

$PT=\sqrt{(\dfrac{4}{3}-0)^2+(0-2)^2}=\dfrac{2}{3}\sqrt{13}$

$PN=\sqrt{(-3-0)^2+(0-2)^2}=\sqrt{13}$

$TA=\dfrac{4}{3}$

$AN=3$

4.

$x^2 + y^2 -8x -6y + 15 = 0$

$2x+2yy'-8-6y'=0$

$y'=\dfrac{x-4}{3-y}$

Point $(5, 0)$

$slope=m_1=\dfrac{5-4}{3-0}=\dfrac{1}{3}$

Tangent line is

$y-0=\dfrac{1}{3}(x-5)$

The equation of the tangent line in slope-intercept form

$y=\dfrac{1}{3}x-\dfrac{5}{3}$

Find the equation of the normal at the point $(5,0)$

$slope=m_2=-\dfrac{1}{m_1}=-3$

$y-0=-3(x-5)$

The equation of the normal line in slope-intercept form

$y=-3x+15$

$TA$   is defined to be the subtangent at $P$. $AN$  is called the subnormal.

The lengths $PT$  and $PN$  are called the tangent and normal.

Points $P, T, N$ and $A$ coincide. Then

$PT=PN=TA=AN=0$