**I already have the answers but I need help understanding how to solve the equa

Question

**I already have the answers but I need help understanding how to solve the equations. Please write out all steps! Thank you so much!** 41. Ellipse (a) Use implicit differentiation to find an equation of the tangent line to the ellipse 1 at (1, 2) (b) Show that the equation of the tangent line to the ellipse xox yoy 1 at Cao, yo) is a2 b2 42. Hyperbola (a) Use implicit differentiation to find an equation of the tangent line to the hyperbola 1 at (3, -2) (b) Show that the equation of the tangent line to the hyperbola 1 at (xo, yo) is yoy

Explanation / Answer

41) a) x2/2 + y2/8 = 1

Taking derivative on both sides of the equation, we get :

2x/2 + 2y/8 dy/dx = 0

=> x + y/4 dy/dx = 0

=> x = -y/4 dy/dx

=> dy/dx = -4x/y

At (1,2) ; dy/dx = -4/2 = -2

Now the equation of tangent line is given by in general:

y - y0 = dy/dx|(x_0,y_0) (x-x0)

y - 2 = -2 (x-1)

=> y-2 = -2x+2

=> y = -2x+4

=> y = -2x+4 is required tangent to the ellipse

(b) The derivative is given by :

2x/a2 + 2y/b2 dy/dx = 0

=> dy/dx = -xb2/ya2

At (x0, y0) ; equation is given by :

y-y0 = -b2x0/a2y0 (x-x0)

=> (y-y0) (a2y0) = -b2x0 (x-x0)

=> ya2y0-a2y02 = -b2xx0 + b2x02

=> ya2y0 + b2xx0 = a2y02 + b2x02

Divide both sides by a2b2, we get :

yy0/b2 + xx0/a2 = y02/b2 + x02/a2

Now since (x0, y0) is on ellipse and so right side equals 1 here

=> yy0/b2 + xx0/a2 = 1 is equation of tangent line to the ellipse

42) a) x2/6 - y2/8 = 1

Taking derivative on both sides we get :

2x/6 - 2y/8 dy/dx = 0

=> x/3-y/4 dy/dx = 0

=> dy/dx = 4x/3y

At (3,-2); dy/dx = 12/-6 = -2

The equation of tangent line is given by :

y+2 = -2 (x-3)

=> y+2 = -2x+6

=> y = -2x+8 is required equation of tangent line to hyperbola

b)

The derivative is given by :

2x/a2 - 2y/b2 dy/dx = 0

=> dy/dx = b2x/a2y

At (x0,y0); the equation is given by :

y-y0 = b2x0/a2y0 (x-x0)

=> (y-y0) (a2y0) = b2x0 (x-x0)

=> ya2y0-a2y02 = b2xx0 - b2x02

=> ya2y0 -b2xx0 = a2y02-b2x02

Dividing both sides by a2b2

=> yy0/b2 - xx0/a2 = y02/b2 - x02/a2

Taking minus sign common through out and cancelling, we get :

xx0/a2 - yy0/b2 = x02/a2 -y0/b2

As (x0,y0) lies on ellipse, hence the right hand side of the eqution is 1

xx0/a2 - yy0/b2 =1 is required equation of tangent line to the hyperbola

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.